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Question

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

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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,

ADB = ADC = 90° (Given, ADBC)
AB = AC (Given)
AD = AD (Common side)

By RHS congruence criteria,
ΔABD ≅ ΔACD

BD = CD (CPCT)

Hence, AD bisects BC.

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

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