Question

In the figure AB=AC and PB= PC prove that
< PAB = < PAC


if you don't know please don't answer​

Answer 1

Answer:

In ∆ PAB and ∆ PAC

AB = AC (given)

PB = PC (given)

AP = PA (common)

∆ PAB ≈ ∆ PAC (SSS congruency)

<PAB = <PAC (c.p.c.t)

Answer 2

Answer:

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