Question

In an Isosceles triangle show that bisector of angle formed between two similar sides is also a altitude, a median and a perpendicular bisector of side of that triangle

Answer 1

In $\Delta ABD$ and $ACD,$
$AB=AC$
$\angle BAD=\angle CAD[\because AD,$ is bisector of $\angle A$]
and $AD=AD$
$\therefore \Delta ADB\cong \Delta ACD$
$\therefore BD=CD$
AD is also a median
and $\angle ADB=\angle ADC={90}^{\circ }$
Thus AD is perpendicular bisector of BC