Download CO-ORDINATE GEOMETRY

GEOMETRY
Similar Triangles
SIMILAR TRIANGLES AND THEIR PROPERTIES
Definition Two triangles are said to be similar if:
(i)
Their corresponding angles are equal, and
(ii)
Their corresponding sides are proportional.
It follows from this definition that two triangles ABC and DEF are similar if:
(i)
A = D, B = E, C = F, and
(ii)
AB BC AC


.
DE EF DF
Note: In the later part of this chapter we shall show that the two conditions given in the above definition are
not independent. In fact, if either of two conditions holds, then the other holds automatically. So, any one of
the two conditions can be used to define similar triangles.
SOME BASIC RESULTS ON PROPORTIONALITY
In this section we shall discuss some basic results on proportionality.
Theorem 1:
Corollary
(Basic proportionality Theorem or Thales Theorem) If a line is drawn parallel to one side of a
triangle intersecting the other two sides, then it divides the two sides in the same ratio.
A
If in a ABC, a line DEBC, intersects AB in D and AC in E, then:
AB AC
AB AC

(i)
(ii)
.

DB EC
AD AE
D
E
Theorem 2:
B
C
(Converse of Basic Proportionality Theorem) If a line divides any two sides of a triangle in
the same ratio, then the line must be parallel to the third side.
Illustration 1: Prove Pythagoras theorem by means of similar triangles.
Solution: Let ABC be a triangle right angled at A. Let AD be perpendicular to BC.
From triangle ADB,
DAB = 90 – B
BAC = 90; DAC = 90 – DAB = 90 – (90 – B) = B
Triangle ADB and ABC are equiangular and hence similar.
A
AB AD AD
AD BD
90



i.e.,

BC AC AC
AC AB
90 B
Cross–multiplying AC2 = DC.BC
Adding (1) and (2)
90 90
AB2 + AC2 = BD.BC + DC.BC = BC(BD + DC)
B
C
D
= BC.BC = BC2 (Since BD + DC = BC)
i.e., sum of the squares on the sides containing the right–angle is equal to square on the
hypotenuse, which is Pythagoras theorem.
http://kishoremaths.in/
Illustration 2: The areas of two similar triangle ABC and PQR and 64cm2 and 121cm2 respectively. If QR = 15.4
cm, find BC.
Solution:
The two triangles ABC and PQR are similar. Hence the ratio of the areas of them is equal to the
ratio of the squares of their corresponding sides.
P
A
ABC BC2


PQR QR2
64
BC2

121 (15.4)2
82
8

 (15.4)2    15.4 
2
11
 11

BC2 = (8  1.4)2
 BC = 8  1.4 = 11.2 cm
BC2=
C Q
B
R
2
Illustration 3: ABCD is a trapezium in which AB || CD and AB = 2CD. Find the ratio of the areas of triangles AOB
and COD.
Solution:
Construction: Join BD and AC,
In two triangles AOB and COD,
BDC = ODC = OBA
OCD = OAB
AOB = COD, (vertically opposite angles.)
 the two triangles are equiangular and hence similar.
COD CD2
CD2
1
( AB = 2CD)




2
2
AOB AB
4
4CD
AOB

4
COD
D
C
O
B
A
Illustration 4: XY || AC and XY divides triangular region ABC into two parts equal in area. Determine
Solution:
XY is drawn parallel to AC. In two triangles BXY and BAC, B is Common
BXY = BAC
BYX = BCA
 ABXY and BAC are equiangular and hence similar.
BXY BX2


BAC BA 2
Now the line XY divides the triangular region into two parts whose areas are equal.
If BAC = 2.
A

BX2
1 BX2
1
BX





2 BA 2
2 BA 2
X
2 BA
1
BA  AX
AX

 1
BA
AB
2

Exercise 1:
AX
.
AB
AX
1
 1

AB
2
2 1
B
Y
C
2
In figure DE  BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
http://kishoremaths.in/
Exercise 2:
Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA
meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB  TC.
Exercise 3:
ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at
L. Prove that
(i)
Exercise 4:
DP
PL
=
DC
BL
(ii)
DL
DP
=
AL
DC
ABCD is a quadrilateral; P, Q, R and S are the points of trisection of sides AB, BC, CD and DA
respectively and are adjacent to A and C; prove that PQRS is a parallelogram.
Exercise 5:
The side BC of a triangle ABC is bisected at D; O is any point in AD. BO and CO produced meet
AC and AB in E and F respectively and AD is produced to X so that D is the mid-point of OX.
Prove that AO : AX = AF : AB and show that FE  BC.
More Results Based on Basic Proportionality Theorem and its Converse
Theorem 3:
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio
of the sides containing the angle.
BD AB

Theorem 4:
In a triangle ABC, if D is a point on BC such that
, prove that AD is the bisector of
DC AC
A.
Theorem 5:
The external bisector of an angle of a triangle divides the opposite side externally in the ratio
of the sides containing the angle.
Exercise 6:
In figure AD is the bisector of BAC. If AB = 10cm,
AC = 14 cm and BC = 6 cm, find BD : DC.
A
1 2
14 cm
10 cm
x
B
Exercise 7:
C
The bisector of interior A of ABC meets BC in D, and the bisector of exterior A meets BC
produced in E. Prove that
Exercise 8:
6-x
D
BD
BE
=
CD
CE
.
ABCD is a quadrilateral in which AB = AD. The bisector of BAC and CAD intersect sides BC
and CD at the points E and F respectively. Prove that EF BD.
Exercise 9:
O is any point inside a triangle ABC. The bisectors of AOB, BOC and COA meets the sides
AB, BC and CA in point D, E and F respectively. Show that AD  BE  CF = DB  EC  FA.
Exercise 10:
AD is a median of ABC. The bisectors of ADB and ADC meets AB and AC in E and F
respectively. Prove that EF  BC.
Theorem 6:
The line drawn from the mid-point of one side of a triangle parallel to another side bisects
the third side.
http://kishoremaths.in/
Theorem 7:
The line joining the mid-points of two sides of a triangle is parallel to the third side and half of
the third side.
Theorem 8:
The diagonals of a trapezium divide each other proportionally.
Theorem 9:
If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.
Theorem 10:
Any line parallel to the parallel sides of a trapezium divides the non-parallel side
proportionally.
Theorem 11:
If three or more parallel lines are intersected by two transversals, prove that the intercepts
made by them on the transversals are proportional.
COROLLARY
If three or more parallel straight lines make equal intercepts on a given transversal,
prove that they will make equal intercepts on any other transversal.
Characteristic Properties of Similarity
We have defined similarity of two triangles. Two triangles are said to be similar iff (i) their corresponding
angles are equal and (ii) their corresponding sides are proportional. Thus, two triangles ABC and DEF are
similar if.
AB BC CA


(i) A = D, B = E, C = F,
(ii)
DE EF FA
Equiangular Triangles
Two triangles are said to be equiangular, if their corresponding angles are equal
Theorem 12: (AAA Similarity) If two triangles are equiangular, then the triangles are similar.
(Corresponding sides are those, which are opposite to equal angles)
D
A
G
B
C
E
H
F
Data: Let ABC, DEF be two triangles, such that A = D; B = E and C = F
AB BC AC


To Prove:
DE EF DF
Proof: Cut off DG = AB and DE and DH = AC from DF, Join GH.
In s ABC and GDH
BAC = GDH
AB = DG
AC = DH
 ABC = GDH
 ABC = DGH
But ABC = DEF (by data)
 DGH = DEF
But there are corresponding angles.
 GH is parallel to EF.
DG DH


DE DF
DG = AB and DH = AC
http://kishoremaths.in/
AB AC

DE DF
Similarly it can be proved that
AB BC

DE EF
AB BC AC



DE EF DF
 If in two triangles ABC and DEF if A = D; B = E and C = F then
AB BC AC


DE EF DF
i.e., if two triangles are equiangular, then their corresponding sides are proportional i.e., if a, b, c are
the length of the sides of the triangle ABC and d, e, f are the lengths of the triangle DEF then
a b c
   k(a cons tant) ,
d e f
Then a = kd; b = ke; c = kf.
The converse of the above theorem is given below.
Theorem 13:
similar.
(SSS Similarity) If the corresponding sides of two triangles are proportional, then they are
D
A
E
B
C
F
G
Data: Let ABC and DEF be two triangles such that
AB BC AC


DE EF DF
To Prove: A = D; B = E; C = F
Construction: On side of EF opposite to D, Make FEG = B and EFG = C
Proof: In triangles ABC and GEF
FEG = B and EFG = C
 EGF = A
 the two triangles ABC and FEG are equiangular and hence their corresponding
AB BC


EG EF
AB BC
But
(By data)

DE EF
 EG = DE.
Similarly it can be proved that FG = DF
In the triangles DEF and GEF,
EF is common
DE = EF
DF = FG
Hence the s DEF and GEF are congruent
DEF = GEF = B (by construction)
and DFE = EFG = C
http://kishoremaths.in/
Triangles ABC and DEF are equangular
Theorem 14: (SAS Similarity) If in two triangles, one pair of corresponding sides are proportional and the
included angles are equal then the two triangles are similar.
D
A
R
C
G
H
E
F
AB AC
Data : Let ABC and DEF be two triangles in which A = D and

DE DF
To Prove: Triangles ABC and DEF are equiangular,
Construction: Along DE, cut off DG = AB and along DF cut off DH = AC, Join GH
Proof: In triangles ABC and DGH
AB = DG
AC = DH
A = D
 the two triangles ABC and DGH are congruent
 DGH = B
AB AC


(Data)
DE DF
DG DH
i.e.,

DE DF
Hence GH is parallel to EF
 DGH = corresponding E
But DGH = B
 B = E
Also A = D (data)
 C = F
Hence the triangle ABC, DEF are equiangular.
AB BC AC


DE EF DF
Theorem 15: If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the
ratio of the corresponding angle bisector segments.
Remarks:
If two triangles ABC and DEF are similar, then
Theorem 16: If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the
ratio of the corresponding attitudes.
Theorem 17: If one angle of a triangle is equal to one angle of another triangle and the bisector of these
equal angles divides the opposite side in the same ratio, prove that the triangles are similar.
Theorem 18: If two sides and a median bisecting one of these sides of a triangle are respectively
proportional to the two sides and the corresponding median of another triangle, then the
triangles are similar.
http://kishoremaths.in/
Theorem 19: If two sides and a median bisecting the third side of a triangle are respectively proportional to
the corresponding sides and the median of another triangle, then the two triangles are similar.
Theorem 20: The ratio of the area of two similar triangles is equal to the ratio of the squares of any two
corresponding sides.
Theorem 21: The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
Theorem 22: The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
Theorem 23: The areas of two similar triangles are in the ratio of the squares of the corresponding angle
bisector segments.
Theorem 24: If the areas of two similar triangles are equal, then the triangles are congruent i.e. equal and
similar triangles are congruent.
Illustration 5: ABCD is a parallelogram. The side CD is bisected at E and BE meets AC at F. Prove
that AF = 2/3 AC.
Solution:
In triangles AFB and EFC AFB = EFC (vertically opposite angles)
FAB = alt. ECF
 third angles are equal. Hence the two triangles are equiangular and hence their
corresponding sides are proportional.
E
C
D
AF AB DC 2EC 2




FC EC DC EC
1
F
AF 2


FC 1
AF
2
2



A
B
AF  FC 2  1 3
AF 2
2
i.e.,
 or AF  AC
AC 3
3
Illustration 6: Two sides of a triangle are 10m and 15m long, and the base is 20m long. If another triangle similar
to the first, has the base of length 32.5m, then find the lengths of the other sides.
Solution:
Exercise 11:
Let k be the ratio of the corresponding sides
32.5 325 13


k=
20
200
8
The length of the other sides of the second triangle are 10k and 15k
13 130
13 195
i.e., 10 
= 16.25m and 15 


 24.375m .
8
8
8
8
Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is
one half the area of the similar triangle ACF described on the diagonal AC as base.
Exercise 12:
Prove that the area of the equilateral triangle described on the side of a square is half the area
of the equilateral triangle described on its diagonal.
http://kishoremaths.in/
Exercise 13:
In figure ABCD is a trapezium in which AB DC
D
C
and AB = 2 DC. Determine the ratio of the areas
of AOB and COD.
o
A
Exercise 14:
In figure PA, QB and RC are each perpendicular to
1 1 1
+ = .
AC. Prove that
x z y
B
P
R
x
Q
A
Exercise 15:
z
C
B
C
In figure, if A = CED, prove that  CAB -  CED.
8
Also, find the value of x.
D
10
x
7
E
2
Exercise 16:
B
9
A
A
In figure, P is the mid-point of BC and Q is the mid-point of AP. If
R
BQ when produced meets AC at R, prove that RS = ⅓ CA.
Q
S
B
Exercise 17:
C
P
D
The diagonal BD of a parallelogram ABCD intersects the segment
C
E
4
AE at the point F, where E is any point on the side BC. Prove that
DF  EF = FB  FA.
1
F
2
3
A
Exercise 18:
B
Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find
the ratio of their corresponding heights.
Pythagoras Theorem
Statement In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the
order two sides.
Converse of Pythagoras Theorem
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the
angle opposite to the first side is a right angle.
http://kishoremaths.in/
Theorem 1:
(RESULT ON OBTUSE TRIANGLE) In the given figure,  ABC
is an obtuse triangle, obtuse-angled at B. If AD  CB, Prove that
AC2 = AB2 + BC2 + 2BC . BD.
A
D
C
B
Remark
In the above theorem BD is known as the projection of AB on BC and the theorem can also be stated as:
In another obtuse triangle, the square of the side opposite to obtuse angle is equal to the sum of the squares
of other two sides plus twice the product of one side and the projection of other on first.
Theorem 2:
(RESULT ON ACUTE TRIANGLE) In the given figure, B of ABC
is an acute angle and AD  BC, prove that
AC2 = AB2 + BC2 – 2 BC . BD
A
B
D
C
Remark
In the above theorem BD is known as the projection of AB on BC and the theorem can also be stated as:
In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of
other two sides minus twice the product of one side and the projection of other on first.
Theorem 3:
Prove that in any triangle, the sum of the squares of any two sides is equal to twice the
square of half of the third side together with twice the square of the median which bisects the
third side.
Theorem 4:
Prove that three times the sum of the squares of the side of a triangle is equal to four times
the sum of the squares of the medians of the triangle.
Illustration 7:
ABC is a right triangle, right angled at B. Let D and E be any points
A
on AB and BC respectively. Prove that AE2 + CD2 = AC2 + DE2
Solution:
Since  ABE is right triangle, right-angled at B.
 AE2 = AB2 + BE2
…..(i)
again,  DBC is right triangle, right-angled at B.
 CD2 = BD2 + BC2
…..(ii)
D
B
E
C
Adding (i) and (ii), we get
AE2 + CD2 = (AB2 + BE2) + (BD2 + BC2)
= (AB2 + BC2) + (BE2 + BD2)
= AC2 + DE2
[Using Pythagoras theorem for  ABC and  DBE,
we have AC2 = AB2 + DE2 BE2 + BD2]
Hence, AE2 + CD2 = AC2 + DE2.
Theorem 5: The perpendicular AD on the base BC of a  ABC intersects BC at D so that DB = 3 CD. Prove
that 2 AB2 = 2 AC2 + BC2.
http://kishoremaths.in/
Illustration 8:
A
ABC is a right triangle, right-angled at C. Let BC = a, CA = b, AB = c
and let p be the length of perpendicular from C on AB, prove that
(i) cp = ab
(ii)
1
1
2
2
p
Solution:
a
+
1
b2
c D
b
p
Let CD  AB. Then, CD = p.
1
1
1
 area of  ABC = (Base  height) =
(AB  CD) = cp.
2
2
2
1
1
Also, area of  ABC = (BC  AC) =
ab
2
2
1
1
cp =
ab  cp = ab
2
2
(ii) Since  ABC is a right triangle, right angled at C.
 AB2 = BC2 + AC2
 c2 = a2 + b2
B
C
a
2
 ab 
2
2
 
 =a +b
p


2 2
ab
a2  b2
1
1
1
1
2
2
1

=
a
+
b

=
 12 = 2 + 2  2 = 2 + 12
a2b 2
b
p2
p
a
a
b
p2
p
Illustration 9:
A
Prove that three times the square of any side of an equilateral
triangle is equal to four times the square of the altitude.
Solution:
Let ABC be an equilateral triangle and let AD  BC.
In  ADC, we have:
AB = AC (given)
B
C
D
B = C
and ADB = ADC
[Each equal to 900]
 ADB  ADC.
So, BD = DC
1
 BD = DC =
BC.
2
Since,  ADB is a right triangle, right-angled at D.
AB2 = AD2 + BD2
1

 AB2 = AD2 +  BC 
2

2
AB
 AB2 = AD2 +
4
Exercise 19:
2
 AB2 = AD2 +

BC2
4
3
AB2 = AD2  3 AB2 = 4 AD2.
4
In figure  AMB   CMD; determine MD in terms of x,
D
B
y and z.
x
M
y
z
A
C
http://kishoremaths.in/
Exercise 20:
D
In figure AB  DC. Prove that (i)  DMU   BMV
U
(ii) DM  BV = BM  DU
M
B
v
A
Exercise 21:
C
Corresponding sides of two triangles are in the ratio 2 : 3. If the area of the smaller triangle in
48 cm2, determine the area of the large triangle.
Exercise 22:
In  ABC, ray AD bisects A and intersects BC in D. If BC = a, AC = b and AB =c, prove that
(i) BD =
ac
(ii) DC =
b+c
ab
b+c
Exercise 23:
In  ABC, C is an obtuse angle. AD  BC and AB2 = AC2 + 3 BC2. Prove that BC = CD.
Exercise 24:
A point D is an the side BC of an equilateral triangle ABC such that DC =
1
4
BC. Prove that
AD2 = 13 CD2.
Exercise 25:
Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square
of half of the third side together with twice the square of the median which bisects the third
side.
Exercise 26:
Prove that in an equilateral triangle, three times the square of a side is equal to four times the
square of its altitudes.
Exercise 27:
In a triangle ABC, N is a point on AC such that BN  AC. If BN2 = AN. NC, prove that
B = 900.
Answers to exercise:
1. x = 4
21. 108 m. sq.
6. BD = 5/2, DC = 7/2
15. x = 6
19. DM = yz/x
http://kishoremaths.in/
ASSIGNMENTS
SUBJECTIVE
LEVEL – I
1.
Show that any point on the bisector of an angle is equidistant from the arms of the angle.
2.
Show that the straight lines which join the extremities of the base of on isosceles triangle to the
middle points of the opposite sides are equal to one another.
3.
In the triangle ABC, the bisectors of exterior angles B and C meet at O. Given
that BAC = 70, ACB = 50. Find BOC.
A
70
B
D
50 C
O
E
4.
If two straight lines are perpendicular to two other straight lines each to each, then show that the
acute angle between the first pair is equal to the acute angle between the second pair.
5.
In a quadrilateral ABCD, if AB = AD and BC = DC, then show that the diagonal AC bisects the
angles, which it joins and that AC is perpendicular to BD.
6.
(i) Show that in a triangle, any two sides are together greater than twice the median, which bisects
the remaining side.
(ii) Show that in any triangle, the sum of the medians is less than the perimeter.
7.
Among all straight line segments drawn from a given point to a given straight line, show that the
perpendicular is the least.
8.
If the opposite sides of a quadrilateral are equal, then show that the figure is a parallelogram.
9.
Show that the line segment joining the meddle points of two sides of a triangle is parallel to the third
side and is equal to half the third side.
10.
Show that the straight line drawn through the middle point of a side of a triangle, parallel to the base,
bisects the remaining side.
http://kishoremaths.in/
LEVEL – II
1.
In ABC, the bisector AX of A intersects BC at X. XL r AB and XM r AC are drawn. Prove that
XM = XL.
2.
PBC and QBC are two isosceles triangle on same base BC but on opposite side of line BC. Show
that the line joining P and Q bisect BC at right angle.
3.
In the given fig. PR = QR. SR = TR. N, M are mid points of SR and TR
respectively. PRT = QRS. Prove that QM = PN.
N
P
S
R
M
Q
4.
T
In the ABC, the sides AB, AC are produced and bisectors of the exterior angle B and C are drawn
to meet in I1. Prove that BI1C = 90 – A/2.
5.
In a quadrilateral ABCD. If A = C and B = D, then prove that the figure in a parallelogram.
http://kishoremaths.in/
OBJECTIVE
LEVEL – I
1.
The area of equilateral triangle with side ‘a’ is
3 a.
(A)
(C)
2.
2a
3
3 2
a . Then its altitude is
4
3a
(B)
2
2a
(D)
3
In equilateral  ABC, AD is altitude. Then 4 AD2 = ……..
(A) 2 BD2
(B) 2DC2
(C) BC2
(D) 3AB2
A
B
3.
In an equilateral  ABC, BC is
(A) 9 DE2
(C) 9 AD2
trisected at D.
(B) 9 BD2
(D) 9 EC2
Then 7 AB2 = ……
A
B
4.
In the figure, P and Q are midpoints of AC
and BC. Then 5 AB2 = …..
(A) 4(AQ2 + BP2)
(B) 4(AC2 + BC2
(C) AP2 + BP2
(D) PC2 + CQ2
In figure, CD  AB CD = P. Then
(A) c
(B)
(C) p
b
(D)
b
E
D
C
A
P
C
5.
C
D
B
Q
A
c
= …..
a
C
b
p
D
P
ab
pc
B
a
C
LEVEL – II
http://kishoremaths.in/
1.
2.
In  ABC, AD, BE & CF are medians. Then
4(AD2 + BE2 + CF2) =
(A) 3(OA2 + OB2 + OC2)
(B) 3(OE2 + OF2 + OD2)
(C) 3(AB2 + BC2 + AC2)
(D) 3(AE2 + AF2 + AD2)
A
F
B
o E
D
In  ABC right angled At C, AD is median. Then AB2 =
(A) AC2 – AD2
(B) AD2 – AC2
(C) 3 AC2 – 4 AD2
(D) 4 AD2 – 3 AC2
A
B
3.
C
In  ABC, BE  AC and CF  AB then BC2 =…..
(A) (AB  BF) + (AC  CE)
(B) (AB  AF) + (AC  AE)
(C) (AB  CF) + (AC  BE)
(D) AB + BC + AC
A
F
E
B
4.
In the figure,  ABC is isosceles. Then AD2 – AC2 =
(A) AD  CD
(B) BC  CD
(C) BD  CD
(D) AB  AD
5.
In  ABC, AB > AC and AD  BC. Then AB2 – AC2 = …..
(A) BD2 – AD2
(B) BD2 – CD2
2
2
(C) BD – AC
(D) AD2 – BC2
C
D
C
A
B
C
D
http://kishoremaths.in/
ANSWERS
SUBJECTIVE
LEVEL – I
1.
7.
10.
2
2
m /n
4 metres
(a  b)2
a2
5.
9.
44.5
CA = 39, CB = 26
LEVEL – II
2.
4.
4
6
3.
5.
9
7
OBJECTIVE
LEVEL – I
1.
D
2.
C
3.
B
4.
A
5.
D
LEVEL – II
1.
C
2.
D
3.
A
4.
C
5.
B
http://kishoremaths.in/