Question:
AD is an altitude of an isosceles ΔABC in which AB = AC.
Show that (i) AD bisects BC, (ii) AD bisects ∠A.
Solution:
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,
AB = AC (Given)
AD = AD (Common side)
ΔABD ≅ ΔACD
Hence, AD bisects BC.
(ii)
Hence, AD bisects ∠A.