Question

The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively,then BD = DC.If this statement is true , enter 1 otherwise enter 0

Answer 1

Given: $△ABC$ is an Isosceles triangle. O is the point of intersection of angle bisectors. $OB$ bisects $\angle B$ and $OC$ bisects $\angle C$
Now, $AB=AC$
$\angle ABC=\angle ACB$ (Isosceles triangle property)
$\frac{1}{2}\angle ABC=\frac{1}{2}\angle ACB$
$\angle OBD=\angle OCD$
In $△OBD$ and $△OCD$,
$\angle ODB=\angle ODC$ (Radius is perpendicular to tangent at the point of contact)
$\angle OBD=\angle OCD$ (From above)
$OD=OD$ (Common)
Hence, $△OBD\cong △OCD$ (ASA rule)
Thus, $BD=CD$
Thus, D bisects BC.