Question

AD is an altitude of an isosceles ΔABC in which AB = AC.
Show that (i) AD bisects BC, (ii) AD bisects ∠A.

Answer 1

Given: AD is an altitude of an isosceles ΔABC in which AB = AC.

To prove: (i) AD bisects BC, (ii) AD bisects ∠A

Proof:
(i) In ΔABD and ΔACD,

$\angle$ADB = $\angle$ADC = 90$°$ (Given, AD$\perp$BC)
AB = AC (Given)
AD = AD (Common side)

$\therefore$ By RHS congruence criteria,
ΔABD ≅ ΔACD

$\therefore$ BD = CD (CPCT)

Hence, AD bisects BC.

(ii)
$\because$ ΔABD ≅ ΔACD [From (i)]
$\therefore$ $\angle$BAD = $\angle$CAD (CPCT)
Hence, AD bisects ∠A.