Question

(3) Two triangles BAC and BDC, right angled at
A and D respectively, are drawn on the same
base BC and on the same side of BC. If AC and
BD intersect at P then complete the following
activity to prove AP x PC=DP X PB.
A
D
P.
B
С please answer​

Answer 1

Answer:

ANSWER

In △APB and △DPC, we have

∠A=∠D=90

∘

and

∠APB=∠DPC [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

△APB∼△DPC

⇒

DP

AP

=

PC

PB

⇒ AP×PC=DP×PB [Hence proved]

I HOPE IT'S HELPFULL

Answer 2

Answer:

In ∆APB & ∆DPC,

∠A =  ∠D = 90°

∠APB = ∠DPC

[ vertically opposite angles]

∆APB  ~ ∆DPC

[By AA Similarity Criterion ]

AP/DP = PB/ PC

[corresponding sides of similar triangles are proportional]

AP × PC = DP × PB

Hence, proved.