Question

19. Prove that in a triangle the difference of any two
sides is less than the third. (Fig. 8.100)

Answer 1

Step-by-step explanation:

Given :-

Triangle ABC and D is the point on AC

Required To Prove :-

Prove that in a triangle the difference of any two

sides is less than the third side .

Construction :-

Construct a point D on AC such that AD = AB

Join B and D.

Proof :-

ABC is a triangle .

D is the point on AC.

AC = AD + DC

=> CD = AC - AD

and we have, AD = AB

=> CD = AC - AB ------------(2)

Now,

In ∆ ABD,

AD = AB ( Construction )

=> <ABD = <ADB -----------(2)

Since the angles opposite to equal sides are equal.

< BDA is an exterior angle formed by extending AD to C .

<BCD is the opposite interior angle.

=> (<BDA) > (<CBD) ------------(3)

In ∆ABD ,

<BDC is an exterior angle and <BDA is an opposite interior angle.

=> (<BDC) > (<BDA) ------------(4)

From (2),(3)&(4)

=> (<BDC) > (<CBD)

=> BC > CD

Since The side opposite to largest angle is longer side.

=> BC > AC - AD

But AD = AB

=> BC > AC - AB

=> AC - AB < BC

The difference of any two sides of triangle is less than the third side.

Hance, Proved.

Used formulae:-

→ The side opposite to largest angle in a triangle is longer side.

→ The angles opposite to equal sides in a triangle are equal.