Question

plz solve this question

Answer 1

in triangle QBP and PBR
<PQB=<PRB
PB=PB
PQ=PR
Hence triangles QBP is congruent to PBR
Then angle QBP is equal to PBR (c.p.c.t )
Then proved that PB bisects angle ABC
Hope this helped you

Answer 2

in ∆BQP and ∆BRP

angle BQP = angle BRP (each 90°)

QP = RP (given)

BP = BP (common hypotenuse)

therefore , by RHS congruence criteria

∆BQP IS CONGRUENT WITH ∆BRP

BY CPCT (corresponding parts of CONGRUENT triangles) => angle QBP = angle PBR
=> BP bisects angle ABC

hope this helps