Question

in fig. PQ=RQ and angle PQS=angle RQS. prove that angle APQ= angle BRQ​

Answer 1

Answer:

In ∆PQS and ∆RQS

=> PQ = RQ ( GIVEN )

=> Angle PQS = Angle BRQ ( GIVEN )

=> SQ = SQ ( COMMON )

by SAS rule of congrency

∆ PQS similarly to ∆ PQS

so,

Angle SPQ = Angle SRQ

Now,

180 - Angle SPQ = 180 - Angle SRQ

Angle APQ = Angle BRQ

H.P

Answer 2

$\huge \mathtt {Question:}$

In the fig. PQ=RQ and < PQS= < RQS. Prove that < APQ= < BRQ.

Given,

PQ = RQ

< PQS = < RQS

To prove:

< APQ = < BRQ

Solution:

PQ = RQ (Given)

< PQS = < RQS (Given)

SQ = SQ (Common side)

Therefore,

∆SQR is congruent to ∆SQP (SAS congruence rule)

< APQ = < BRQ (Linear pairs of congruent parts of congruent triangles)

Hence proven.