Question

If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

Answer 1

Proved , that point lies on the bisector of that angle.

Given :

• ED & DF are congruent i.e. ED=DF

• ED&DF are perpendicular to AB &

BC respectively.

i.e. < DEB=<DFB= 90°

Proof

• In triangle EDB & triangle FDB

• < DEB=<DFB ( Each 90°)

• ED = DF ( Given)

• DB = DB ( Common)

• triangle EDB is congruent to

triangle FDB

• <EBD=<FBD ( CPCT)

•This means, BD bisects < EBC

& D lies on BD

hence proved

Answer 2

Proved below.

Step-by-step explanation:

Given:

Let P be a point within ∠ABC such that PM = PN.

To prove:

P lies on the bisector of ∠ABC i.e., ∠1 = ∠2.

Proof:

In ΔPMB and ΔPNB, 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