Question

Question 2 (ii)
AD is an altitude of an isosceles triangles ABC in which AB = AC.
Show that AD bisects $\angle A$.

Answer 1

$In\Delta BADand\Delta CAD$,
$\angle ADB=\angle ADC$ (Each measure ${90}^{\circ }$, as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
$\therefore \Delta BAD\cong \Delta CAD$ (By RHS Congruence rule)
By CPCT,
$\angle BAD=\angle CAD$
Hence, AD bisects $\angle A$