PROVE THAT ANY POINT ON THE ANGLE BISECTOR OF AN ANGLE IS EQUIDISTANT FROM ITS ARMS. 

Hi!
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.
 
Cheers!

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thank you ,sir!!!!!!!!!!!!!!!!!11

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are Thnku SIR

:)

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