Question

please solve this sum fast​

Answer 1

triangle apb and bpc

aqp =pqb (each 90)

bc=bc (commen side)

qpb=pbc (alternate angle)

∆APB =~∆PBC by sas Congre

pa=pb =1/2 ac

Answer 2

Answer:

given: a ∆abc such that pa= pc

construction: pq is drawn // to bc

to prove: pa= pb= ½ ac

proof:-

since pq is // to bc, q is the mid point of ab.

then by theorem ar∆ aqp = ar∆qpb

then, ∆aqp is congruent to ∆ qpb

ap=pb ----------1

but ac = ap+pc

ac= 2ap (since ap= pc)

ap=½ ac

then equation 1 can be presented as

ap= pb= ½ ac

hope this help