Question 2 (i) AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
AD bisets BC
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Solution
In ΔBAD and ΔCAD, ∠ADB=∠ADC(Each 90∘ AD is an altitude)
AB = AC (Given)
AD = AD (Common) ∴ΔBAD≅ΔCAD(By RHS Congruence rule) ⇒BD = CD (By CPCT) Hence, AD bisects BC.