6.13 Theorems on Triangles and circles:
6.13.1 Basic
proportionality theorem
You must have heard about pyramids in
They used the concept of ‘similarity’ of polygons
to measure the heights and distances which are otherwise difficult to find.
Similarity concept is extensively used in vehicle
manufacturing, space technology and many other engineering fields
We have figures of two pentagons
on the right side What do you observe? Don’t you notice that they are
identical except for their sizes? Actually
the figure on extreme left side is 50% smaller than the figure on its right side. When
you measure the length of the sides of these figures, You will notice that they
are in the same ratio. AB/ PQ = BC/ QR
= CD/ RS = ……= 1/ 2 You
will also notice that the angles at each of the corresponding vertices are
same. _{}EAB = _{}TPQ, _{}ABC = _{}PQR, _{}BCD = _{}QRS , . . . 

Two polygons having same
number of sides are similar (denoted by the symbol ‘’) if and only if:
1. The corresponding
angles of two polygons are equal
AND
2. The corresponding
sides of two polygons are proportional (i.e. ratios of their sides are equal)
Notice that
regular polygons of same number of sides are always similar.
If two rectangles are to be similar then
the ratio of their lengths = ratio of their breadths
as in the figure given below
Let
us observe the following figures; Don’t they appear of same shape? _{}ABC = _{}DEF, _{}BAC =_{}EDF,
_{}ACB = _{}DFE The
_{}ABC and _{}DEF are similar triangles and we represent this by _{}ABC  _{}DEF . 


Congruent Figures 
Similar Figures 
1 
Figures
have same shape and same size. 
Figures have same shape and sizes are in same proportion 
2 
Corresponding
sides are equal 
Ratios of corresponding sides are same 
3 
Congruent triangles are similar. 
Similar triangles need not be congruent. 
In 6.3.3 we have observed that it is not possible
to construct unique triangle given three angles.
If two triangles have two pairs of corresponding
angles equal, then triangles are similar (This is called ‘AAA (Angle, Angle,
Angle) Postulate
on similarity’.
Note: The above postulate is also called ‘AA (Angle, Angle)
Postulate on similarity’.(The
reason is that, if two angles are equal in a triangle then
third angle has to be equal
because sum of all angles in a triangle is 180^{0})
6.13.1. Basic Proportionality Theorem (Thale’s theorem)  (BPT) A straight line drawn
parallel to a side of a triangle, divides the other 2 sides
proportionately.
Data :In _{}ABC, X is a point on AB and the line XY is BC,
To Prove: AX/BX = AY/CY.
Step 
Statement 
Reason 


Consider _{}ABC and _{}AXY 

1 
_{}AXY = _{}ABC 
Corresponding
angles(XYBC) 

2 
_{}AYX = _{}ACB 
Corresponding
angles(XYBC) 

3 
_{}XAY = _{}BAC 
Common
angle 

4 
_{}ABC  _{}AXY 
AAA
Postulate on similarity 

5 
AB/AX
= AC/AY 
Corresponding
sides are proportional 

6 
(AX+BX)/AX
= (YC+AY)/YC 
Spilt
AB and AC 

7 
1+BX/AX =
1+AY/YC 
Simplification 

8 
BX/AX
= AY/YC 
Subtract
1 from both sides of Step 5 

9 
AX/BX =AY/CY 
Reciprocal
of 6 
6.13.1 Problem 1: In the above figure if XY  BC, prove that AB/BX=AC/YC
Step 
Statement 
Reason 

1 
AX/XB=AY/YC 
BPT 

2 
1+(AX/XB)=1+(AY/YC) 
Add 1 to both sides 

3 
(XB+AX)/XB=(YC+AY)/YC 
X and Y are points on
AB and AC 

4 
AB/XB=
AC/YC 

Conversely, If a straight line divides the two sides of a triangle
proportionately then that line has to be parallel to the third side.
(Note: Proof is not provided for this (The
proof is by
logical reasoning).
Note: Basic
proportionality theorem’s concept can be used to divide a given line in to
different ratios (Refer 6.1 Problem 11)
6.13.1. Corollary to BPT:
If a line is drawn
parallel to a side of a triangle, then the sides of the new triangle formed are
proportional to the sides of the given triangle
(New triangle is ‘similar’ to the given triangle)
Data: XY is parallel to BC
To prove: AX/AB=AY/AC=XY/BC
Construction: Draw a line parallel to AC from X and
let that line meet BC at Z
Proof:
Step 
Statement 
Reason 

1 
BZ/BC=BX/BA 
BPT
since XZ  AC (Construction) 

2 
XY=ZC, YC =XZ 
By
construction, XZCY is a parallelogram and hence opposite sides are equal 

3 
AX/AB
= AY/AC 
BPT
since XY  BC (Given) 

4 
LHS
=AX/AB = (ABBX)/AB 
X
is a point on AB 

5 
=
1(BX/AB) 


6 
=
1(BZ/BC) 
From
(1) 

7 
=
(BCBZ)/BC 
Z
is a point on BC 

8 
= ZC /BC 


9 
=XY/BC= AY/AC 
From(2) 
6.13.1 Problem 2: Show that in a trapezium,
the line joining the midpoints of non parallel sides is parallel to the
parallel sides
Data: BCED is a trapezium; X and Y are midpoints of
DB and CE respectively
To prove: XY  DE
Construction: Extend DB and EC to meet at A (since
DB and EC are non parallel lines, they have to meet at a point when extended)
Step 
Statement 
Reason 

1 
AB/BD
= AC/CE 
BPT
(DE  BC) 

2 
(AXBX)/BD
= (AYCY)/CE 
X
and Y are points on DA and EA 

3 
AX/BD
– BX/BD = AY/CE –CY/CE 
X
and Y are mid points so BD=2BX=2DX,CE=2YE=2CY 

4 
AX/2DX
1/2 =AY/2YE 1/2 
From
3 

5 
AX/DX
=AY/YE 
Add
1/2 to both sides in 4 and then
multiply both sides by 2 

6 
XY
 DE 
Converse
of BPT 
6.13.1. Theorem 1: If two triangles are
equiangular then their corresponding sides are proportional
Data: In _{}ABC and _{}DEF
_{}BAC = _{}EDF
_{}ABC =_{}DEF
_{}ACB =_{}DFE
To Prove: AB/DE = BC/EF= AC/DF
Construction: Mark X on AB and Y on AC such that
AX=DE and AY=DF. Join XY
Proof:
Step 
Statement 
Reason 

1 
_{}BAC = _{}EDF 
Given 

2 
AX=DE,AY=DF 
Construction 

3 
_{}AXY and _{}DEF are congruent 
SAS Postulate 

4 
_{}AXY=_{}DEF 
Congruent triangle property 

5 
= _{}ABC 
Given 

6 
XYBC 
Corresponding angles _{}AXY and _{}ABC are equal 

7 
AB/AX=AC/AY=BC/XY 
Corollary to BPT 

8 
AB/DE=AC/DF=BC/EF 
From(2) 
6.13.1. Converse of Theorem 1: If the corresponding sides of two triangles are proportional
then the triangles are
equiangular
Data: In _{}ABC and _{}DEF
AB/DE=AC/DF=BC/EF
To Prove:
_{}BAC = _{}EDF
_{}ABC =_{}DEF
_{}ACB =_{}DFE
Construction: Mark X on AB and Y on AC such that
AX=DE and AY=DF. Join XY
Step 
Statement 
Reason 

1 
AB/DE=AC/DF=BC/EF 
Given 

2 
AX =DE,AY=DF 
Construction 

3 
AB/AX=AC/AY 
Step 1 and 2(Substitution) 

4 
XYBC 
Converse of BPT 

5 
_{}AXY=_{}ABC, _{}AYX=_{}BCA 
Corresponding angles are equal 

6 
_{}AXY=_{}DEF, _{}AYX=_{}DFE 
By
construction _{}AXY, _{}DEF are congruent 

7 
_{}ABC=_{}DEF, _{}BCA=_{}DFE 
Step 5, 6(Substitution) 

8 
_{}BAC =_{}EDF 
When two angles are equal third
angle has to be equal 
6.13.1 Problem 3: ABC is a triangle in which AB=2cm, BC=3cm and CA =
4cm. DEF is another triangle whose largest side is 6cm. If DEF is similar to
ABC,
find the relationship
between perimeters of these two triangles.
Solution:
The
perimeter of _{}ABC= 2+3+4 =9cm Let
the largest side in _{}DEF be DF=6cm Since
triangles are similar we have DE/AB
= EF/BC =DF/AC Since
DF=6 and CA = 4, DF/AC =6/4=1.5 _{}DE = 1.5*AB = 1.5*2 = 3cm and EF =1.5*BC=1.5*3=4.5cm The
perimeter of _{}DEF = 6+3+4.5 =13.5cm which is one and half times the perimeter of _{}ABC. 

Thus, the perimeters of similar triangles are in same proportion
as their corresponding sides
6.13.1 Problem 4: You must have heard about Meharuli
Pillar near Qutubminar. It is an iron pillar installed
by Chandragupta II who ruled northern
Assume that on a visit to the place, you are
standing 9feet 2inches away from this pillar and you notice that the shadow
cast by the pillar is
2 feet 8 inches. If your
height is 5’4”. Find the height of the pillar.
Solution:
For
simplicity let us convert all measures to inches. Let
the pillar be represented by AB. You are at D, 110 inches (9’2”) away from
the pillar. Your
height is represented by DP = 64” (5’4”). Shadow cast by pillar (DC) is 32
inches. _{}BAC =_{}PDC = 90^{0} Since
AB  DP, _{}ABP = _{}DPC _{}ACB is a common angle to both _{}BAC and _{}PDC. Therefore
these two triangles are equiangular and hence their corresponding sides are
proportional(AAA Postulate) Hence
AB/PD = AC/DC. Note that AC =AD+DC = 110+32= 142 _{}AB = AC*PD/DC = 142*64/32=284’’ The
height of iron pillar = 284 inches = 23 feet 8inches. You
will notice from history books that this answer is correct. 

6.13.1 Problem 5: _{}ABC is a right angled triangle at B and D is any point on
Solution:
_{}ABC =_{}DEA = 90^{0}. Since A is the common vertex of
triangles ABC and DAE, _{}BAC = _{}DAE. Since
two angles of these triangles are same, third angle also has to be same (_{}BCA=_{}ADE). Therefore
_{}ABC and _{}DEA are equiangular (similar) triangles. Therefore
by theorem, corresponding sides are proportional. The
corresponding sides are: AC,
AD (hypotenuse) BC,
DE (opposite side of _{}A) AB,
AE (opposite to _{}BCA and _{}ADE) _{} AE/AB = DE/BC=AD/AC _{}AE = AB*AD/AC = 16*4/24 = 16/6 =2.67m 

6.13.1 Problem 6: In the trapezium ABCD,
ADBC and the diagonals intersect at O. Prove that OB/OC = OD/OA
Solution:
Since
ADBC, _{}BCO = _{}OAD, _{}CBO=_{}ODA (alternate angles) Since
the lines AC and BD intersect at O, _{}COB =_{}AOD (vertically opposite angles) _{} The triangles BCO
and DOA are equiangular. Hence
by theorem the corresponding sides are proportional The
corresponding sides are OC,
OA (opposite to angles CBO and ODA) _{} By theorem OB/OD =
OC/OA i.e.
_{}OB/OC = OD/OA 

6.13.1 Theorem 2: Areas of similar triangles are proportional to the
squares of the corresponding sides
Data: Let ABC and DEF be similar triangles, in
which (BC,EF), (AB,DE) and (AC,DF) are pairs of corresponding sides.
To Prove: Area of _{}ABC/Area of _{}DEF = BC^{2}/EF^{2}= AB^{2}/DE^{2}
=AC^{2}/DF^{2}
Construction: Draw
Step 
Statement 
Reason 

1 
Area of _{}ABC= BC*AL/2 
Area of triangle = base*height/2 

2 
Area of _{}DEF = EF*DM/2 
Area of triangle = base*height/2 

3 
(Area of _{}ABC)/ (Area of _{}DEF)=(BC/EF)*(AL/DM) 


4 
_{}ABL =_{}DEM 
Triangles ABC and DEF are
similar (Given) 

5 
_{}ALB =_{}DME= 90^{0} 
Construction 

6 
_{}BAL =_{}EDM 
Since 2 angles are equal, third
angle is also equal 

7 
AL/DM = AB/DE =BL/EM 
Corresponding sides are
proportional in similar triangles 

8 
But AB/DE =BC/EF 
Triangles ABC and DEF are
similar (Given) 

9 
_{}AL/DM=BC/EF 
From (7) and (8) 

10 
(Area of _{}ABC)/ (Area of _{}DEF)=(BC/EF)*(BC/EF)= BC^{2}/EF^{2} 
Substitute (9) in (3) 

11 
Similarly we can prove other 2
equalities 
6.13.1 Corollary : If the areas of two similar
triangles are equal then they are congruent
Proof :
Let
ABC and DEF be two similar triangles with BC and EF as base _{}By above theorem (Area
of _{}ABC)/ (Area of _{}DEF) = BC^{2}/EF^{2} Since
the areas are same it follows that BC=EF, Similarly
we can prove that other corresponding sides are also equal Hence
by SSS Postulate they are congruent. 

6.13.1 Problem 7: Prove that the area of similar triangles have the
same ratio of the squares of corresponding altitudes
Solution:
In
the above theorem, in step 7, we had proved that triangles ALB and DME are
equiangular. Because
of this, the corresponding sides are proportional (Theorem 1) _{} AB/DE = AL/DM _{}(AB/DE)^{2 }= (AL/DM)^{2} Since
the triangles ABC and DEF are similar, by the above theorem Area of_{}ABC /Area of _{}DEF = BC^{2}/EF^{2} = AB^{2}/DE^{2} = AL^{2}/DM^{2 } =
squares of the ratio of the altitudes 

6.13.1
Problem 8: In _{}ABC, BE is Perpendicular to AC and CF is perpendicular to AB.
BE and CF intersect at O.
Prove that Area of_{}BOF / Area of _{}COE= BF^{2}/CE^{2}
^{ }
Step 
Statement 
Reason 

1 
_{}BFO =_{}CEO = 90^{0} 
Given 

2 
_{}BOF =_{}COE 
Opposite angle 

3 
_{}FBO =_{}OCE 
Since 2 angles are equal, third
angle has to be equal 

4 
_{}BFO  _{}CEO 
Triangles are equiangular 

5 
Area of_{} BFO /Area of _{}CEO = BF^{2}/CE^{2} 
By theorem 
6.13.1 Problem 9: In the
adjoining figure ABC and DBC are two triangles on the same base. Prove that Area
of_{} ABC /Area of _{}DEF = AO/DO
Construction: Draw Perpendicular to base BC from A
and D to cut the line BC at E and F.
Step 
Statement 
Reason 

1 
_{}AOB =_{}DOF 
Vertically opposite angle 

2 
_{}AEO =_{}DFO= 90^{0} 
Construction 

3 
_{}AEO _{}DFO 
AAA postulate on similarity 

4 
AE/DF = AO/DO 
Corresponding sides are
proportional 

5 
Area of_{} ABC /Area of _{}DEF =
{(1/2)BC*AE}/{(1/2)BC*DF}=AE/DF 
Formula for area 

6 
Area of_{} ABC /Area of _{}DEF=AO/DO 
Step 4 
6.13.1 Problem 10: Prove that
internal bisector of an angle of a triangle divides the opposite side in the
ratio of the sides containing the angle.
(This is also called vertical angle
theorem)
Construction: Draw a line from C parallel to AD to
meet the extended BA at E(DACE)
To Prove: AB/AC = BD/DC
Step 
Statement 
Reason 

1 
DACE 
Construction 

2 
BD/DC = BA/AE 
BPT on _{}BCE 

3 
_{}BAD =_{}AEC 
Corresponding angles DACE 

4 
_{}DAC =_{}ACE 
Alternate angles DACE 

5 
_{}BAD =_{}DAC 
AD is bisector of _{}BAC 

6 
_{}AEC =_{}ACE 
Step 4,5 

7 
AE=AC 
Step 6 (_{}CAE is isosceles) 

8 
BD/DC = BA/AC 
Step 2 
Note: Converse Of the above problem is that, if a line segment drawn from
the vertex of an angle of a triangle to its opposite side divides the line
in the ratio of the sides containing
the angle, then the line segment bisects the angle.
6.13.1 Problem 11: In the
Adjoining figure diagonal BD of a quadrilateral ABCD bisects _{}B and _{}D Prove that AB*CD =AD*BC
Hint : 1.
_{}ADB = _{}CDB, _{}ABD = _{}CBD( Given that BD bisects
_{}B and _{}D) 2.
BY AAA postulate on similarity _{}ADB _{}CDB 3.
Hence corresponding sides are proportional: AB/BC =AD/CD _{} AB*CD =AD*BC 

6.13.1 Problem 12: What is the
height of the intersecting point joined by strings tied to the base and top of
poles of height 15 and 10 ‘Mola’s( A unit of measurement)? ( Lilavati Shloka
162) .
Step 
Statement 
Reason 

1 
OP/CD = BP/BD 
OP
 CD ,
BPT 

2 
OP/AB=PD/BD 
OP
 AB , BPT 

3 
OP/CD +OP/AB = (BP/BD)+ (PD/BD) 
Add
(1),(2)


4 
OP(AB+CD)/(AB*CD) = (PD+BP)/BD =1 
(PD+BP) =BD 

5 
OP
= (AB*CD)/ (AB+CD) = 15*10/25 = 6 


Note : we can find BP
or PD given PD or BP 
6.13.1 Problem 12: O Lilavati, Tell me the distance between the
toy and the ‘Deepa’
( Candle stand) and the height of ‘Deepa’ if:
the height of toy is 12” and the height
of its shadow is 8” in its original place
height of the shadow becomes
12” if toy is moved forward by 2 ‘Hasta’ ( 1 Hasta = 24”)
( Lilavati Shloka
244). Bhaskara also gives formulae to solve these types of problems.
Solution : In the figure
AB is the Deepa, CL and EM are the
initial and final positions of the toy whose height is 12”.
CD is the shadow of the toy at C and the length of
the shadow is 8”.
New position of the toy is at E (CE = 48”( 2*24))
EF is the shadow of the toy at E its length is 12”.
Step 
Statement 
Reason 

1 
AB/LC = BD/CD 
LC
 AB ,
BPT 

2 
AB/ME =BF/EF 
ME
 AB ,
BPT 

3 
BD/CD = BF/EF 
LC=ME=12, From (1),(2) 

4 
(BC+CD)/CD = (BC+CD+DE+EF)/12 
BD = (BC+CD), BF =(BC+CD+DE+EF) , EF=12 

5 
(BC/8)+1
= (BC/12)+5 
CD=8, (CD+DE+EF) = 8+40+12 =60 

6 
BC(1/81/12)
= 4 


7 
BC{(64)/48}
=4 _{} BC =96 _{} BD = 104 
BD
= BC+CD = 96+8 

8 
AB
= (BD*LC)/CD= (104*12)/8 = 156 
From
(1) 
6.13.2 Pythagoras Theorem:
In a right angled triangle, the square of
hypotenuse is equal to sum of squares of other 2 sides.
There are around 50 proofs for this theorem and we
use one of the simplest proof here.
It is generally represented as
AC^{2} = AB^{2}+BC^{2} when
right angled at B
Data: In triangle _{}ABC = 90^{0}
To Prove: AC^{2} = AB^{2}+BC^{2}
Construction: Draw BD perpendicular to hypotenuse
AC
Proof:
Step 
Statement 
Reason 

1 
_{}ABC= 90^{0} 
Given 

2 
_{}BDA= 90^{0} 
Construction 

3 
_{}BAC =_{}BAD 
Common angle 

4 
_{}_{}ABD =_{}BCD 
Since 2 angles are equal, third
angle has to be equal 

5 
_{}ABC  _{}ADB 
Triangles are equiangular 

6 
_{}AB/AD = AC/AB 
Ratios of sides are in
proportion(6.13.1 Theorem 1) (side opposite to _{}C and hypotenuse) 

7 
_{} AB^{2} =
AC*AD 
Simplification 

8 
_{}ABC  _{}CDB 
Repeat the steps 1 to 5 for _{}ABC and _{}CDB 

9 
BC/CD=AC/BC 
Ratios of sides are in
proportion(6.13.1 Theorem 1) (side opposite to _{}A and hypotenuse) 

10 
_{} BC^{2} =
AC*CD 
Simplification 

11 
_{} AB^{2} + BC^{2}
=AC*AD+AC*CD =AC(AD+CD)=AC*AC= AC^{2} 
Add (7) and (10) and note D is a
point on AC 
Have you noticed the following?
1. 25 =5^{2}
= 3^{2}+4^{2 }= 9+16
2. 100 = 10^{2}
= 8^{2}+6^{2 } =64+36
The triplets like (3,4,5),
(6,8,10), (8,15,17), etc are called Pythagorean triplets
Note : It was Baudhayana (600BC), an
Indian mathematician who first found
this relationship in a right
angled triangle !
Converse of Pythagoras Theorem: If the square of one side is
equal to sum of squares of other two sides then these two sides(AB,BC)
contain a right angle
(If AC^{2} = AB^{2}+BC^{2}
then _{}B= 90^{0})
Proof is provided by constructing another right
angled triangle and then proving that both are congruent (By using SSS
postulate).
6.13.2 Problem 1: A school has
a compound wall of 400 feet. A painter was asked to paint the wall. The painter
charges Rs 80 per sq. ft to paint the wall.
The painter has a ladder of length 10 feet. When
the ladder rests against the wall its foot is 6 feet away from the compound
wall. Find out how much the
school pays the painter to
paint the compound wall.
Solution:
If
we find the area of the compound wall, we can multiply this area by the rate,
to know howmuch the painter is to be paid for his
work. We
know AC^{2} = AB^{2}+BC^{2} Substituting
values for AC=10 and BC=6 we get 10^{2}
= 6^{2}+AB^{2 } By
transposition we get AB^{2}=10036
=64 _{} AB = Height of the
wall = 8 feet Area
of the compound wall = 400*8 = 3200 sq ft. Since
the painter charges Rs 80 per sqft. Total amount
payable is Rs 2,56,000 (=3200*80). 

6.13.2 Problem 2: You are selected from your school to participate in
chess a game being conducted in another school. You cycled to reach that
school as follows:
From your house you cycled 8 km to the north and
then 5 km to east and then 4 km to the north.
How many km did you peddle to reach the other school?
Solution:
Draw
a figure to represent AB= 8km, BC=5km and CD=4 km. The route you cycled is
ABCD. We
are required to find DA. Draw
a line DE parallel to CB to meet the extended line AB at E. Since
DE is parallel to BC and BE is parallel to CD, BCDE is a rectangle _{}DE= 5 and AE
= AB+BE = 8+4 =12 Note
that DEA is a right angled triangle _{} AD^{2} = ED^{2}+EA^{2
}= 5^{2}+12^{2 }=
25+144 = 169 = 13^{2} _{} AD= 13km 

6.13.2 Problem 3: AD is the altitude
through A in the _{}ABC and DB:CD = 3:1. Prove that BC^{2}
= 2(AB^{2}AC^{2})
Step 
Statement 
Reason 

1 
BD^{2}
= AB^{2}AD^{2} 
Pythagoras
theorem on _{}ABD 

2 
AC^{2}
= AD^{2}+CD^{2} 
Pythagoras
theorem on _{}ADC 

3 
_{} AD^{2} = AC^{2}CD^{2} 
Step
2 

4 
_{} BD^{2} = AB^{2}(AC^{2}
CD^{2}) = AB^{2}AC^{2} +CD^{2} 
(1) and (3) 

5 
DB/CD
= 3 _{} BC = BD+CD =4CD 
Given 

6 
_{} DB=3CD _{} BD^{2}=9CD^{2}

Step
5 

7 
_{} 9CD^{2} = AB^{2}AC^{2}
+CD^{2} 
(6)
and (4) 

8 
_{} 8CD^{2} = AB^{2}AC^{2} 
Transposition 

9 
_{} 16CD^{2} =
2(AB^{2}AC^{2}) 
Multiply
8 by 2 

11 
_{}BC^{2}= 2(AB^{2}AC^{2}) 
Step
5 
6.13.2 Problem 4: A Bamboo pole
of height 32 ’Mola’(Unit
of measurement : say feet) broke due to
heavy wind and fell on the ground. If the pole’s top touched the ground at 16 ‘Mola’ away from
its base, find the point at which it broke.(Lilavati Sholka 150)
Solution:
AB
is the pole of height 32 feet.Let the pole break at
D to touch the ground at C BC
= 16 feet. By Pythagoras theorem DC^{2} = DB^{2}+BC^{2 }= DB^{2}+16^{2 }= DB^{2}+256 Let BD = x and DC = y ^{} So
we have two equations to solve x+y =32 à(1) And y^{2}
= x^{2}+256 à(2) From
(1) we get y = 32x By
substituting this value of y in (2) we get (32  x)^{2 }= x^{2}+256 LHS
= 32^{2}+ x^{2 } 64x (expand (ab)^{2}= a^{2}+
b^{2}2ab) = 1024 + x^{2 } 64x Since
LHS= RHS _{}1024 + x^{2 } 64x = x^{2}+256 _{} 768 = 64x (Transposition) _{} x = 12 _{} The pole breaks at
12feet from the bottom. Verification: since x=12, y =
3212 = 20 We observe that 400 = 20^{2} = 144+256 = 12^{2}+16^{2} This
is as per Pythagoras theorem. 

6.13.2 Problem 5: A monkey
climbed down from a tree of height 100 ‘Mola’(
A unit of measurement like feet) and went to a pond 200 ‘Mola’ away. Another monkey jumped up a little distance and
then jumped down diagonally and reached the same pond. If both monkeys
travelled the same distance, tell me the height to which the second monkey
jumped?( Lilavati : Shloka 157)
BD represents the tree of height 100 Mola(say
Feet) C is the pond 200 Mola(Feet) away. Let x be the
height to which the monkey jumped from the top
of tree BD. Distance travelled.
Solution:
BD
represents the tree of height 100 Mola(say Feet). C is the pond 200 Mola(Feet) away. Let
x be the height to which the monkey jumped from the top of tree BD.(DA=x) Distance
travelled by the first monkey = DB+BC = 300 Distance
travelled by the second monkey = DA+AC = x+AC It
is given that distance travelled by
both monkeys are same _{} x+AC=300
_{} AC= 300x _{} AC^{2} ^{ }= (300x)^{2 }=90000600x+
x^{2}  (1) By Pythagoras theorem AC^{2} = BA^{2}+BC^{2 }= (100+x)^{2}+200^{2 }= 10000+200x+x^{2}+40000 (2) From (1) and (2) 10000+200x+x^{2}+40000 =90000600x+
x^{2} _{} 800x = 40000 _{} x=50 Verification : AC^{2} = (100+50)^{2}+ 200^{2}
= 22500+40000 = 625000 = 250^{2} 100+200 = 50+ 250 

6.13.2 Problem 6: A peacock
is sitting on the top of a pillar of height 9 ‘Mola’(feet).
An ‘ant hill’ exists at the base of the
pillar. To hide itself from the attack of peacock, a snake is approaching the
‘ant hill’ from
a distance three times the height of the pillar. Seeing the snake approaching
the ‘ant hill’, peacock flies down to catch the snake. If
both snake and peacock travel at the same speed, find the distance from ‘ant hill’
at which peacock catches the snake. ( Lilavati : Shloka 152)
Solution:
AB
is the pillar of 9 feet. Peacock is sitting at A. D is the position of snake
at a distance of 27 feet from the pillar. Peacock follows the path (AC) of a triangle
to catch the snake. Let
C be the place where peacock catches the snake. Let BC=x. _{} CD=27x Since
both peacock and snake travel at same speed AC =CD _{} AC=27x _{} AC^{2} ^{ }= (27x)^{2 }=72954x+ x^{2}^{
} (1) From
Pythagoras theorem AC^{2} = BA^{2}+BC^{2 }= (9)^{2}+x^{2 }= 81+x^{2} (2) From
(1) and (2) 72954x+ x^{2} =81+x^{2}
_{} 72981 = 54x _{} x= 648/54=12 Verification :: AC^{2} = (2712)^{2} =
9^{2} + 12^{2}
( _{} 225 = 81+144) 

6.13.2 Problem 5: A circus group has put up a tent around a central pole
of height of 11 meters. At 12 meters distance, from the foot of the central
pole,
the circus group has put up few poles of height 6meters on
the ground in a circular fashion around the central pole. The poles are
strengthened by ropes
tied between the top of
central pole and top of the other poles around it. Find the length of the rope
required to tie the poles at the top.
Solution:
The
erection of poles can be represented as in the adjoining figure. BD
is the central pole of height = 11 Meters AC
is one of the supporting poles of height 6 meters, put up on the ground, around the central
pole. The
distance between the central pole(BD) and the supporting pole(AC) on the
ground (AB) = 12 meters We
are required to find DC. From C, draw a line parallel to AB to cut BD at E. ABEC
is a rectangle with CE=12 and BE=AC=6 _{}ED=5 CED
is a right angled triangle Hence
by Pythagoras theorem DC^{2}
= CE^{2}+ED^{2}= 12^{2}+5^{2}=144+25 =169 =
13^{2} _{}CD = 13 meters Therefore
the length of the rope required to tie these 2 poles at the top = 13 meters. 

6.13.2 Problem 6: In the adjoining figure _{}ACB = 90^{0} AB=c, BC=a, AC= b, _{}CDB = 90^{0}and CD =p
Prove that 1/p^{2} = 1/a^{2} + 1/b^{2}_{}
Step 
Statement 
Reason 

1 
AB^{2} = AC^{2}+BC^{2} 
Pythagoras theorem for _{}ACB 

2 
c^{2} = b^{2}+a^{2} 
Substitution 

3 
(1/2)AB*CD = (1/2)cp 
Area of _{}ACB with base = AB 

4 
(1/2)AC*BC = (1/2)ba 
Area of _{}ACB with base = AC 

5 
(1/2)cp = (1/2)ba 
Both areas are same 

6 
c = ab/p 
Step 5 

7 
a^{2}b^{2}/ p^{2}
= b^{2}+a^{2} 
Substitute value of c in step 2 

8 
1/ p^{2}= (b^{2}+a^{2})/
a^{2}b^{2} 
Divide both sides by a^{2}b^{2} 

9 
1/ p^{2}= 1/a^{2}+1/b^{2} 

6.13 Summary of learning
No 
Points
to remember 
1 
A straight line drawn parallel
to a side of a triangle divides the other 2 sides proportionately (BPT) 
2 
If two triangles are equiangular
then their corresponding sides are proportional 
3 
Areas of similar triangles are
proportional to the squares of their corresponding sides 
4 
In a right angled triangle, the
square of hypotenuse is equal to sum of squares of other 2 sides 
Additional points:
6.13.1 Theorem 3: Areas of triangles having a common vertex and
having bases along the same line, are proportional to the
lengths of their bases.
In the adjoining figure triangles _{}ABP,_{}ABC and _{}ABD have the same vertex A and their bases are on the same
line BD.
We are required to prove that
Area of ABP/Area of ABC = BP/BC, Area of ABC/ Area
of ABD = BC/BD
Proof:
Since
all the triangles have the same vertex and their bases are on the same
straight line, the heights of all the triangles are same. We
know that Area of a triangle = ½*Base*Height. _{}Area of ABP/Area of ABC = (1/2*BP*height)/(1/2*BC*height) =
BP/BC Similarly,
Area of ABC/ Area of ABD = BC/BD 

6.13.1 Theorem 4: A perpendicular drawn from the vertex of the right
angle of a right angled triangle divides the triangle into two similar
triangles and
they are also similar to
the given triangle.
Given: In the right angled _{} ABC, _{}ABC = 90^{0}, BD_{}AC
To prove: _{}ABC  _{}ADB  _{}BDC
Step 
Statement 
Reason 


Consider _{}ADB and _{}ABC 

1 
_{}ADB = _{}ABC = 90^{0} 
BD_{}AC 

2 
_{}BAD = _{}BAC 
Common


3 
_{}ABC  _{}ADB 
AAA
Postulate 


Consider _{}BDC and _{}ABC 

4 
_{}BDC = _{}ABC = 90^{0} 
BD_{}AC 

5 
_{}BCD = _{}BCA 
Common


6 
_{}ABC  _{}BDC 
AAA
Postulate 




7 
_{}ABC  _{}ADB _{}BDC 
Steps
3 and 6 
Note: In the above diagram prove that
BD^{2} = AD*DC
Since _{}ADB  _{}BDC, AD/BD =BD/DC (Corresponding sides of similar triangles are
proportional)
_{} AD*DC = BD^{2}
6.13.1 Theorem 5 (30^{0}60^{0}90^{0}
triangle theorem) : If the angles of a triangle are
30^{0}, 60^{0} and 90^{0} then the side opposite to 30^{0}
is half the
hypotenuse and the side opposite
to 60^{0} is _{}/2 times the hypotenuse.
Given: In _{} ABD, _{}ABD = 60^{0}, _{}BAD = 30^{0}, _{}ADB = 90^{0}
To prove: BD = AB/2, AD = _{}/2 * AB
Construction: Produce BD such that
DC = BD. Join AC.
Step 
Statement 
Reason 


Consider _{}ADB and _{}ADC 

1 
_{}ADB = _{}ADC = 90^{0} 
Given
that _{}ADB = 90^{0} and Construction 

2 
BD = DC 
Construction 

3 
AD is common 


4 
_{}ADB  _{}ADC 
SAS Postulate 

5 
_{}BAD = _{}CAD 
Corresponding angles of similar
triangles are equal 

6 
_{}BAC = 60^{0} 
Step 5 and it is given that _{}BAD = 30^{0} 

7 
AB = AC = BC 
ABC is an equilateral triangle
(all angles = 60^{0}) 

8 
BD = DC = AB/2^{} 
Step
2, 7 

9 
AB^{2} = AD^{2}+BD^{2} 
Pythagoras theorem 

10 
AB^{2} = AD^{2}+(AB/2)^{2} 
Step 8 

11 
AD^{2 }= AB^{2}(AB/2)^{2
}= (3AB^{2})/4 
Simplification 

12 
AD = _{}/2*AB 

6.13.3
Postulates on similarity:
In addition to AAA postulate on
similarity
discussed in the beginning of this section, we have three more postulates on
similarity.
2.
‘Two triangles are similar if two of their sides are
proportional and their included angles are same’ This
statement is called ‘SAS (Side, Angle, Side) Postulate
on Similarity’. In
the adjoining figure AB/DE = BC/EF and their included angle _{}ABC = _{}DEF. There
fore _{}ABC  _{}DEF. 

3.
‘Two triangles are similar if the sides of one
triangle are proportional to the corresponding sides of another triangle’.
This
statement is called ‘SSS (Side, Side, Side) Postulate on Similarity’. In
the adjoining figure DE/AB = EF/BC = DF/AC There
fore _{}ABC  _{}DEF. 

Table: Postulates for similarity of triangles:
Proportional 
Equal 
Similarity Postulate 

Side 
Side 
Side 
Angle 
Angle 
Angle 

 
 
 
y 
y 
y 
AAA 
Y 
Y 
 
Y(Included) 
 
 
SAS 
Y 
y 
y 
 

 
SSS 
6.13.3 Intercept Theorem: If three or more lines
are intercepted by two transversals, the intercepts made by them on the transversals
are
proportional. (Note this is a variation of the theorem 6.8.7)
Given: Transversal p makes
intercepts on three lines, l, m and n.
(I.e. l  m  n) q is another
transversal which makes intercepts DE and EF
To prove: AB/BC=DE/EF
Construction:
Steps 
Statement 
Reason 


Consider _{}ABS and _{}BCT 

1 
_{}ABS = _{}BCT, _{}BAS = _{}CBT, 
Corresponding angles (It is given that line l  line m) 

2 
_{}ABS  _{}BCT 
AA Postulate On similarity 

3 
AB/BC=AS/BT 
Corresponding sides are proportional 

4 
ASED is a parallelogram 
ASDE(construction), ADBE(Given) 

5 
AS=DE 
Sides of parallelogram 

6 
BTFE is a parallelogram 
BTEF(construction), BECF(Given) 

7 
BT=EF 
Sides of parallelogram 

8 
AB/BC = AS/BT = DE/EF 
Step 3,5 and 7 
6.13.3 Problem 1: In the
adjacent figure the medians IB and JC meet at G. Prove that BG = 2GI.
Hint: Since
BI and CJ are medians, by definition of a median BJ=JA and CI=IA. By
mid point theorem (6.7.7) 2IJ
= BC and IJ  BC Using
the fact that IJ  BC and AAA postulate on similarity prove that _{}GIJ and _{}GBC are similar _{} IJ/BC = GI/BG Since
2IJ = BC, it follows that GI/BG =1/2 i.e.
2GI = BG 

6.13.3 Problem 2: As shown In the
figure given below, M is the mid point of BC of parallelogram ABCD. DM intersects
the diagonal AC at P and
it meets AB produced to E. Prove that PE = 2PD
Given: ABCD is a parallelogram; M
is mid point of BC
To prove: PE = 2PD
Step 
Statement 
Reason 


Consider _{}MCD and _{}MBE 

1 
CM = MB 
Given 

2 
_{}DMC = _{}BME^{} 
Vertically
opposite angles 

3 
_{}DCM = _{}MBE 
Alternate angles ABCD 

4 
_{}MCD _{}_{}MBE 
ASA Postulate 

5 
DM = ME 
Corresponding sides of congruent
triangles are equal 


Consider _{}ADP and _{}MCP 

6 
_{}APD = _{}MPC ^{} 
Vertically
opposite angles 

7 
_{}ADP = _{}PMC 
Alternate angles ADBC 

8 
_{}DAP = _{}PCM 
Alternate angles ADBC 

9 
_{}ADP  _{}MCP 
AAA Postulate 

10 
AD/CM = PD/PM 
Corresponding sides of similar
triangles are proportional (6.13.1 Theorem 1) 

11 
AD/CM = 2^{} 
AD=BC=2CM
(Given) 

12 
PD/PM = 2 
Steps
10 and 11 

13 
PE = PM+ME = PM+DM = PM+DP+PM = 2PM+PD 
ME=DM
as per step 5 

14 
PE = PD+PD = 2PD 
2PM=PD(step 12) 
6.13.3 Problem 3: In the given figure
ABC, DEBC and AD:DB = 3:5. Calculate the ratio of :
(i)AD/AB, AE/AC, DE/BC
(ii) Area of _{}DEO/Area of _{}EOC
(iii) Area of _{}DBE/Area of _{}DBO
(iv) Area of _{}ADE/trapezium DECB
Note:
1. _{}DOE and _{}EOC have bases on the same line (DC) and same vertex E. Hence
their areas are proportional to their bases (=DO/OC).
(Refer 6.13.1 Theorem 3)
2. _{}DBE and _{}DBO have bases on the same line (BE) and same vertex D. Hence
their areas are proportional to their bases (=BE/BO).
3. _{}ABC and _{}ADE are similar triangles and hence their areas are
proportional to the squares of their corresponding sides (=BC^{2}/DE^{2}).
(Refer 6.13.1 Theorem 2)
Steps 
Statement 
Reason 

1 
AB/AD = AC/AE = BC/DE 
DEBC
and BPT Theorem(6.13.1) 

2 
AB/AD = (AD+BD)/AD = 1+BD/AD = 1+ 5/3 = 8/3 
Given
AD:DB = 3:5 

3 
_{} 
DEBC
and Alternate angles 

4 
_{} 
AAA
postulate on similarity 

5 
DE/BC = DO/OC 
corresponding
sides are proportional 

6 
DE/BC = OE/OB 
Follow
above steps to prove this. 

7 
DE/BC = AD/AB = 3/8 
Step
1, Step 2 

8 
Area
of _{}DEO/Area of _{}EOC= DO/OC 
Note 1 

9 
=3/8 
Step 5,7 

10 
Area
of _{}DBE/Area of _{}DBO = BE/BO 
Note 2 

11 
=(BO+OE)/BO
= 1+OE/BO = 1+(3/8)
= 11/8 
Step 6 

12 
Area of trapezium DECB = Area of _{}ABC – Area of _{}ADE 


13 
Area of trapezium DECB/ Area of _{}ADE = (Area of _{}ABC Area of _{}ADE) / Area of _{}ADE = Area of _{}ABC/ Area of _{}ADE1 = (BC/DE)^{2}1 = (64/9)1 = 55/9 
Note 3 
6.13.2 Problem 7: Find the area of an isosceles triangle whose equal
sides is ‘a’ units and base is ‘b’ units
Solution:
In the adjoining figure let
AC=BC=a and AB= b We know AD=DB =b/2
(perpendicular from vertex C bisects the base (AB)) By Pythagoras theorem AC^{2 }=
AD^{2}+DC^{2} DC^{2 }= AC^{2 }
AD^{2}= a^{2}(b/2)^{2}= (4a^{2}b^{2})/4 i.e. _{}Area of _{}ADC = (1/2)*base*height = (1/2)*(b/2)*SQRT(4a^{2}b^{2})/2
= 1/8*b*SQRT(4a^{2}b^{2}) _{}Area of _{}ABC = 2*Area of _{}ADC = ¼*b*SQRT (4a^{2}b^{2}) 

6.13.2 Problem 8: Find the area of an equilateral triangle whose
equal sides are ‘a’ .
Solution:
Since all sides are equal in an equilateral
triangle we can substitute b=a in the formula
arrived in the above problem (6.13.2
Problem 7).
Note that _{}DCB = 30^{0}, _{}DBC= 60^{0}, AD=DB=a/2
_{}Area of an equilateral triangle = 1/4 *a* SQRT (4a^{2}a^{2}) = (_{}/4)a^{2}^{}
Using
Pythagoras theorem we can prove that DC = _{} (a/2)
We may observe that in triangle CDB the angles are
in the ratio of 30:60:90 and the corresponding sides are in the ratio of (½):(_{}/2):1 or 1:_{}:2
6.13.2 Problem 9: The sum of squares of 2 sides of a triangle is
equal to twice the square of half the third side plus
twice the square of the median which bisects the third side.
(This is also
called as Apollonius theorem)
Given: ABC is a triangle, AD is
the median
To prove: AB^{2} + AC^{2} = 2BD^{2}+2AD^{2}
Construction: Draw AE _{}BC
Hint: Apply Pythagoras theorem to three triangles (AEB, AED, AEC). Then substitute and simplify the
terms and also use the given data that BD=DC. 

In section 6.4 we studied many
concepts such as Incenter, Circumcenter, Orthocenter and Centroid. We shall
prove their special properties here.
6.13.4 Concurrency theorems:
6.13.4 Theorem 1: Prove that angular bisectors of a
triangle are concurrent.
Given: In _{}ABC, the bisectors of _{}ABC and _{}BCA meet at
To prove: AI bisects _{}BAC.
Construction: Draw ID_{}BC, IE_{}AC and IF_{}AB
Step 
Statement 
Reason 

1 
ID=IF ^{} 
_{}BDI _{}_{}BFI (ASA Postulate with BI as common side) 

2 
ID=IE^{} 
_{}CDI _{}_{}CEI (ASA Postulate with CI as common side) 

3 
IE=IF 
Steps 1,2 

4 
_{}AFI _{}_{}AEI 
RHS Postulate (hypotenuse AI is
common ) 

5 
_{}FAI=_{}EAI 
Corresponding angles are equal
(Step 4) 

6 
AI
bisects _{}BAC 
Step 5 
Thus angular bisectors are
concurrent at I (Incenter).
6.13.4 Theorem 2: Prove that perpendicular
bisectors of a triangle are concurrent.
Given: In _{} ABP, perpendicular
bisectors SD of BP and SE of AP meet at S and SF is Perpendicular to AB.
To prove: SF is perpendicular bisector
of AB.
Construction: Join SA, SB and SP.
Step 
Statement 
Reason 

1 
SB=SP^{} 
_{}BDS _{}_{}PDS (Given BD=DP) and SAS Postulate with SD as common side 

2 
SP=SA^{} 
_{}PES _{}_{}AES (Given PE=AE) and SAS Postulate with SE as common side 

3 
SB=SA 
Steps
1,2 

4 
_{}AFS _{}_{}BFS 
Step3,
(Given _{}AFS = _{}BFS = 90^{0}) and RHS Postulate with SF as common
side 

5 
AF=BF 
Corresponding
sides of congruent triangles are equal(Step 4) 

6 
SF bisects AB 
Step
5 
Thus perpendicular bisectors are
concurrent at S (Circumcenter).
6.13.4 Theorem 3: Prove that altitudes of a
triangle are concurrent.
Given: In _{}ABC, altitudes AD from vertex A and BE from vertex B meet at
To prove: CF is perpendicular to
AB.
Construction: Through A, B and C
draw parallel lines to the bases BC, CA and AB respectively forming _{}PQR.
Step 
Statement 
Reason 

1 
BCQA is a parallelogram 
By
construction BACQ, BCAQ 

2 
BC=AQ 
Step 1 

3 
BCAR is a parallelogram 
By
construction RABC, RBAC 

4 
RA=BC 
Step 3 

5 
AQ=RA (A is mid point of RQ) 
Steps
2 and 4 

6 
AD is perpendicular to RQ 
Given
: AD _{}BC and by construction BCRQ 

7 
AD is perpendicular bisector of
RQ 
Steps
5 and 6 

8 
BE is perpendicular bisector of
RP 
Repeat
steps 1 to 7 with respect to parallelograms RBCA, ABPC and BE_{}RP 

9 
Perpendicular bisectors AD and
BE of _{}PQR meet at O 
Steps
7 and 8 

10 
CF is perpendicular bisector of
PQ and it passes through O 
Previous
theorem (6.13.1 Theorem 2) in view of Step 9 

11 
CF_{}AB 
By
construction PQBA 
Thus Altitudes are concurrent at O
(Orthocenter).
6.13.4 Theorem 4: Prove that medians of a triangle
are concurrent and their division is in the ratio of 2:1.
Given: In _{}ABC, the medians BE and CF meet at G.
AG produced to meet BC at D.
To prove: BD=DC and AG:GD=BG:GE=CG:GF=2:1
Construction: Produce AD to K such
that AG=GK. Join BK and CK.
Step 
Statement 
Reason 


Consider _{}ABK 

1 
BF=FA 
Given 

2 
GK=AG 
By construction 

3 
FGBK,
FG=(1/2)BK 
Mid Point theorem 


Consider _{}ACK 

4 
EGCK,
EG=(1/2)CK 
Mid Point theorem(Similar to
steps 1,2) 



5 
GCBK 
Step 3 

6 
BGKC 
Step 4 

7 
BGCK
is a parallelogram 
Steps 5 and 6 

8 
The
diagonals BC and GK bisect each other 
Step 7 (Property of
parallelogram) 

9 
BD=DC 
Step 8 

10 
GD
= DK = (1/2)GK 
Step 8 

11 
=
(1/2)AG 
Step 2 

12 
2GD=AG 
Steps 10, 11 
Similarly we can prove that 2GE=BG
and 2GF=GC.
Note: In case of an equilateral triangle, Centroid
(G),Incenter (I),Circumcenter(C, S) and
Orthocenter (O)are same 1. BD=DC, CE=EA, BF=FA (G is the point where the
three medians meet, G=G) 2. AD_{}BC, BE_{}AC, 3. AD, BE, CF are perpendicular bisectors (G is
the point where the three perpendicular bisectors meet, G=S) 4.
_{}BAD =_{}CAD, _{}CBE=_{}ABE, _{}BCF=_{}ACF (G is the point where the three angular bisectors meet,
G=I) 

6.13.5
Relationship between similarity and sizetransformation.
Introduction: You must have seen models of houses, flats,
buildings, townships…They are all scaled down versions of actual constructions.
These
miniature versions give us a
bird’s eye view of actual constructions. These models are ‘reduction’s of
actuals. Similarly enlarged models are used
when the actual
constructions to be built are very small.
Let us study the adjacent figure. ABC is a triangle and P is an
external point. Points A_{1}, B_{1}^{
}and C_{1}^{ }are on the lines PA, PB and PC
respectively such that PA_{1} = (1/2)PA, PB_{1}
= (1/2)PB, PC_{1}^{ }= (1/2)PC and Points A_{2}, B_{2}^{
}and C_{2}^{ }are on the lines PA, PB and PC
respectively such that PA_{2} = 2PA, PB_{2}
= 2PB, PC_{2} = 2PC By joining A_{1}, B_{1},
C_{1}^{ }and A_{2}, B_{2}, C_{2}^{
}we get A_{1}B_{1}C_{1} and A_{2}B_{2}C_{2}
as two other triangles. We observe the following properties
about the triangles ABC, A_{1}B_{1}C_{1} and A_{2}B_{2}^{
}C_{2}:
(We can also prove the above
observations) We define A_{2}B_{2}C_{2}^{
}as the triangle ‘enlarged’ by a ‘scale factor of 2’ with respect to P. Here P is
called as the ‘center of enlargement’ We define A_{1}B_{1}C_{1}
as the triangle ‘reduced’ by a ‘scale factor of 1/2’ with respect to P. Here P
is called as the ‘center of reduction’ The type of enlargement or
reduction discussed above is called ‘Size
Transformation’. Size transformation is not
restricted to triangles and they are applicable for all types of geometrical
figures. Properties of size
transformations:
(i)
k > 1 then the transformation is enlargement (ii)
k < 1 then the transformation is reduction (iii)
k = 1 then there is no transformation(same as original) 

6.13.5 Problem 1: In the following figure, triangles APQ and AMN
are the images of the triangle ABC under enlargement and reduction respectively
with A as center.
Given that MN=4cm, AB=9cm, BC=8cm,
AC=6cm and CQ=3cm,
1. Name the images of B in each
case
2. Find the scaling factor in each
case
3. Find AN, AM, BP and PQ
Solution:
Step 
Statement 
Reason 

1 
_{}ANM = _{}ABC = _{}APQ 
MNBCPQ
(Alternate angles) 

2 
The images of B are N
and P 
Step 1 

3 
ANM is reduction APQ is enlargement 


4 
MN/BC = 4/8 = ½ (scaling factor) for ANM 
MN = 4 and BC = 8 

5 
AQ/AC = 9/6 = 1.5 (scaling factor) for APQ 
AC = 6 and CQ = 3 

6 
1/2 = AN/AB = AN/9 à AN = 4.5cm 


7 
1/2 = AM/AC = AM/6 à AM = 3cm 


8 
1.5 = PQ/BC = PQ/8 à PQ = 12cm 


9 
1.5 = AP/AB = AP/9 à AP = 13.5cm 


10 
13.5 = AP = AB+BP = 9+BP à BP = 4.5cm 
