PROVE THAT ANY POINT ON THE ANGLE BISECTOR OF AN ANGLE IS EQUIDISTANT FROM ITS ARMS.
Here is the answer to your question.
EF is the bisector of ∠BOD and ∠COA. R is a point on EF, RP⊥AB and RQ⊥CD.
In ∆ORP and ∆ORQ,
OR = OR (Common)
∠RPO = ∠RQO (90°)
∠POR = ∠QOR (OR is the bisector of ∠POQ)
∴ ∆ORP ≅ ∆ORQ (AAS congruence criterion)
⇒ PR = RQ (C.P.C.T)
Thus, the perpendiculars drawn from any point on the angle bisector of an angle, to the arms of the angle, are equal.