Question

Two right triangles ABC &DBC are drawn on the same side of BC. If AC & DB intersects at P, prove that AC*PC=BP*PD​

Answer 1

Answer:

hey here is your proof

pls mark it as brainliest

so here we go

Step-by-step explanation:

here consider triangle bap and triangle pdc,

so angle bap=pdc (each 90)

angle apb=dpc (vertically opposite angle)

thus triangle bap similar to triangle pdc by A.A Test of similarity

so ap/dp==bp/pc (c.s.s.t)

ie ap×pc=bp×dp

ie apxpc=bp×pd

thus proved

hope it helped u a lot