Question

Prove that the lengths of the perpendicular from any point on the bisector of an angle to the sides are equal​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let P be a point within ∠ABC such that PM=PN. We have to prove that P lies on the bisector of ∠ABC ie., ∠1=∠2.

In Δ  

′

sPMB and PNB, we have  

PM=PN       [Given]

BP=BP       [common]

∠PNB=∠PMB [Right angle]

So, by RHS congruence criterion, we have  

∴ΔPBM≅ΔPNB

⇒∠1=∠2          [∵∠B=∠C]

⇒P lies on the bisector of ∠ABC