Question

d
In the adjoining figure Bo 1 Ac. CQ LAB, A-P-C,
A-Q-B then prove that​

Answer 1

Answer:

1REF. Image.

Given AB=AC

and AP=AQ

Thus

AB-AP=AC-AQ

[BP=CA ] [from figure ]

now InΔBCP & ΔBCQ

BP = CQ

∠c=∠c [common]

and BC=BC [common]

∴ΔBCP≃ΔBCQ [SAS congruency]

now

[BQ=CP] [corresponding parts of congruent triangles]

solution