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

Question

Philoid · Accepted Answer

Given: A Δ ABC in which BO is bisector of ∠ABC

Also, we have PD ⊥ AB and PE ⊥ BC

To Prove: PD = PE

Proof:

In Δ PBD and Δ PBE

PB = PB [common]

∠ PBD = ∠ PBE [ given]

∠ PDB = ∠ PEB = 90° [Given]

Thus, Δ PBD ≅ Δ PBE [Angle – Angle – Side]

∴ PD = PE

Hence Proved.