Question

in the figure ABC and DBC are two right triangles. prove that AP × PC = PD.​

Answer 1

Answer:

hey here is your answer

so pls mark it as brainliest

pls refer the reference diagram uploaded above

for better understanding

Step-by-step explanation:

To Prove=AP×PC=BP×PD

now considering triangle apb and triangle dpc

so here angle bap=angle pdc=90 (Each 90)

also angle apb=angle dpc because they are vertically opposite angles

thus triangle apb similar to triangle dpc by A.A Test of similarity

so thus then

AP/BP=PD/PC (c.s.s.t)

thus ie AP×PC=PD×BP

ie AP×PC=PD×BP

Thus proved

Answer 2

In ∆ BAP & CDP

$\sf\angle BAP = \angle PDC = 90°$

${\sf\angle BPA = \angle CPD ~~~~----( vertically~~ opposite~~ angle )}$

$\sf {∆BAP \sim ∆ DPC ~~~~~~~~~----( by \: AA \: Similarity )}$

$\sf \dfrac{AP}{DP} = \dfrac{BP }{ PC }$

$\bf AP × PC = BP × DC_{_{_{_{ Proved }}}}$