ATOMIC ENERGY EDUCATION SOCIETYDISTANCE EDUCATION PROGRAMME

CHAPTER : 7

TRIANGLESModule : 4/4

In the previous module we learned about congruence criteria of triangles SAS, ASA (AAS), SSS and RHS.

Also we have discussed properties of triangles i.e. Angles opposite to equal sides of an isosceles triangle are equal and the sides opposite to equal angles of a triangle are equal.

Now in this module we will learn inequalities of a triangle.

Before learning inequalities we will revise the concepts learned in previous module through examples.

1. In fig. AC = BC, ∠DCA = ∠ECB and ∠DBC = ∠EAC. Prove that △DBC ≅△ EAC.

Solution: In △ACE, △BCD

∠EAC = ∠DBC (given)

AC = BC (given)

∠DCA = ∠ECB (given)

add ∠DCE on both sides

∠DCA + ∠DCE = ∠ECB + ∠DCE

∠ACE = ∠BCD

△EAC ≅△DBC (ASA)

A C B

D E

2. In the given fig. △ ABC is an isosceles triangle in which AB = AC and side BA is produced to D such that AD = AB. Show that ∠BCD is right angle.

Solution. △ABC is an isosceles triangle

AB = AC

∠ABC = ∠ACB (angles opp. to equal sides)

Let ∠ABC = ∠ACB = x (1)

In △ACD,

AC = AD (Since AB = AC and AB = AD by producing)

∠ACD = ∠ADC = y (2) (angles opp. to equal sides)

A

B C

D

x xy

y

In △BCD,

∠B + ∠BCD + ∠D = 180° (ASP)

∠x + ∠x + ∠y + ∠y = 180° {From (1) and (2)}

2∠x + 2∠y = 180°

2(∠x + ∠y) = 180°

∠x + ∠y = 180°∕2 = 90°

∠x + ∠y = 90°

∴ BCA is a right triangle.

Now we will learn inequalities of a triangle.

1. If two sides of a triangle are unequal, the angle opposite to the longer side is larger ( or greater).

In △ABC, if AB > AC then

∠C > ∠B

A

B C

(2) If two angles of a triangle are not equal then

the side opposite to the larger angle is

greater (larger) P

In PQR

Q > R then

PR > PQ Q R

(3) The sum of any two sides of a triangle is greater than the third side .

A

AB + AC > BC

AB + BC > AC

BC + AC > AB

B C

Now we will discuss some examples to apply the inequalities of the triangle.

Ex : (1) In the given figure of ABC , A = 200

C = 400 . If the bisector of the angle B meets

AC at X. Prove that BC > XC.

Solution: A 20 X

40

ABC = 1800 – 200 – 400 = 1200 B C

(Angle sum Property)

ABX = CBX = ½ X 1200 = 600

BXC = 1800 – 600 – 400 = 800

In BXC, 800 > 600 BC > XC

(2) In the given fig. ABC is a triangle with B = 350 , C = 650 and bisector of BAC meets BC in X. Arrange AX, BX and CX in descending order.

Solution: In ABC

B + C + BAC = 1800

350 + 650 + BAC = 1800

BAC = 1800 - 1000 = 800

AX is bisector of BAC

BAX = CAX = ½ X 800 = 400

In AXC , 650 > 400 AX > CX (1)

In ABX, 400 > 350 BX > AX (2)

From (1) & (2) BX > AX > CX

A

350 650

B X C

(3) If O is a point with in ABC. Show that OA + OB + OC > ½ ( AB + BC + CA)

Solution :

In OAB OA + OB > AB → (1)

In OBC OB + OC > BC → (2)

In OCA OC + OA > AC → (3)

(1) + (2) + (3)

2( OA + OB + OC) > AB + BC + AC

OA + OB + OC > ½ ( AB + BC + AC)

A

O

B C

(1) Prove that the sum of any two sides of a triangle is greater than twice the length of the median drawn to the third side.

(2) In the figure given here, if AD is the bisector of BAC, then prove that

(i) AB > BD (ii) AC > CD

(3)In the given figure PR > PQ and PS bisects QPR. Prove that

PSR > PSQ.

A

→median

B D C

A

B D C

P

Q S R

THANK YOU

PREPARED BY

K SRIDHAR, AECS – 2, KALPAKKAM