Question

Prove that a triangle AC is isosceles, if:

(i) altitude AD bisects angle $\angle$ BAC, or

(ii) bisector of angle BAC is perpendicular to base BC.

Answer 1

1. In ΔABC, let the altitude AD bisects ∠BAC.

Then we have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ AB = AC (cpct)

Hence, ΔABC is an isosceles.

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

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ AB = AC (cpct)

Hence, ΔABC is an isosceles.