AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
Solution and Explanation
(i) In ∆BAD and ∆CAD,
∠ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∠∆BAD ≅ ∠∆CAD (By RHS Congruence rule)
BD = CD (By CPCT)
Hence, AD bisects BC.
(ii) Also, by CPCT,
∠BAD = ∠CAD
Hence, ∠AD bisects A.
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