Question

pls answer this question correctly​

Answer 1

Answer:

given:AB=AC

AP=AQ

• to prove :-(i)∆CAQ=~∆BAP

AP=AQ

<A=<A

AC=AB

SO,. ∆CAQ=~∆BAP

HENCE, CQ=BP ( by cpct)

• (ii)∆BQC=~∆CPB

<Q=<P ( due to same base)

BQ=CP

BC=BC

SO,. ∆BQC=~∆CPB

HENCE, CQ=BP (by cpct)

proved

VARSHA PARIHAR.

Answer 2

Step-by-step explanation:

i) In Triangle CAQ and Triangle BAP

1. AC=BP (Given)
2. AQ=AP (Given)
3. Angle A is common

Therefore, Triangle CAQ ≅Triangle BAP (By SAS Rule)

Therefore, BP=CP (By CPCT)

ii) In Triangle BQC and Triangle CPB

Since, AB=AC------1

AP=AQ------2

Subtracting 1&2-

AB-AP=AC-AQ

Therefore, BQ=CP

1. BQ=CP (Proved Above)
2. BP=CP (Proved Above)
3. BC is common.

Therefore, Triangle BQC≅Triangle CPB (By SSS Rule)

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