Question

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

Answer 1

Given that in ΔABC, AB = AC
and AD ⊥ BC
i) Now in ΔABD and ΔACD
AB = AC (given)
∠ADB = ADC (given AD ⊥ BC)
AD = AD (common)
$\therefore$ ΔABD $\cong$ ΔACD ($\because$ RHS congruence)
$\Rightarrow$ BD = CD (CPCT)
$\Rightarrow$ AD, bisects BC.
ii) Also ∠BAD = ∠CAD (CPCT of ΔABD $\cong$ ΔACD )
$\therefore$ AD bisects ∠A.