Question

4. In the given figure, line segment AB is bisected by line 1. If 2 PQA = 2 PQB, show that P is equidistant from A and B. ​

Answer 1

Answer:

In triangle PAQ and triangle PQB

P Q = Common

A Q= B Q (Data,)

angle PQA= PQB (DATA,)

triangle PAQ congruent to triangle

PBQ (SAS)

hence PA= PB

Answer 2

$In Triangle \: \: : PAQ and Triangle PQB$

$PQ = COMMON$

$AQ = BQ ( BY SSS )$

$angle \: 2PQA = \: 2PQB \: (BY SSS)$

$triangle \: \: PAQ \: \\ Congruent \: to \: \: triangle$

$PQB = BY (SSS)$

$PA = PB$

$A = B$

$Hence \: it \: is \: Proved$