Question

19. Prove that a triangle ABC is isosceles, if :
(i) altitude AD bisects angle ZBAC, or
(ii) bisector of angle BAC is perpendicular to
base BC.​

Answer 1

Answer:

Step-by-step explanation:

(we have drawn an triangle ABC and AD is the altitude)

Given : A triangle ABC  in which BD = DC

to prove : AB = AC or ABC is an isosceles triangle .

 

 Proof:

 

In triangle ABD and triangle ADC

AD = AD ( common)

∠ADB = ∠ADC (90° each)

BD = DC (given)

so Δ ABD ≡ ΔACD( by RHS)

So AB = AC (by cpct)

∴ Δ ABC is an isosceles triangle.

 

Answer 2

Step-by-step explanation:

19. [ii.]

In Δ ABC, the bisector of ∠ BAC is perpendicular to the base BC. We have to prove that the ΔABC is isosceles.

In ΔABD & ΔADC:

∠BAD = ∠CAD (AD is the bisector of ∠BAC)

AD = AD (Common)

∠ABD = ∠ADC (Each 90⁰ )

ΔABD ≅ ΔADC (A.A.S)

AB = AC (C.P.C.T.C)

Hence, ΔABC is an isosceles.