Question

Question 2 (i)
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
AD bisets BC

Answer 1

In $\Delta$BAD and $\Delta$CAD,
$\angle ADB=\angle ADC$ (Each ${90}^{\circ }$ AD is an altitude)
AB = AC (Given)
AD = AD (Common)
$\therefore \Delta BAD\cong \Delta CAD$ (By RHS Congruence rule)
$\Rightarrow$ BD = CD (By CPCT)
Hence, AD bisects BC.