Question

In figure, BO bisects $\angle ABC$ of $△ABC$. P is any point on BO. Prove that the perpendicular drawn from P to BA and BC are equal.

Answer 1

In the given figure

$BO$ is the angle bisector of $\angle ABC$

Then $\angle DBP=\angle DBE=\frac{1}{2}\angle ABC$

Perpendicular drawn from $P$ to $AB$ and $BC$

In $△PBD$ and $△PBE$

$\therefore \angle PDB=\angle PEB={90}^o$

$BO=BO$ ...... [common side]

$\angle DBP=\angle BDE$

$\therefore △BDP\cong △BEP$ ....... [By ASA rule]

$\therefore PD=PE$

Then, the perpendicular draw from P to BA and BC are equal.