Question

In figure, two chords AB and CD of a circle intersect each other at the point P outside the circle. prove that 1)triangle PAC ~ triangle PDB 2) PA. PB = PC. PD​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\bf\huge\underline{Solution}$

We have two chords AB and CD when produced meet outside the circle at P.

(i) Since in cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.

Therefore, ∠PAC = ∠PDB ⠀⠀⠀ .....(1)

and ∠PCA = ∠PBD ⠀⠀⠀⠀⠀⠀⠀ .....(2)

From (1) and (2) and using AA similarity, we have

∆PAC ~ ∆PDB

(ii) Since, ∆PAC~∆PDB[As proved above]

Therefore, their corresponding sides are proportional.

=> $\dfrac{PA}{PD}$ = $\dfrac{PC}{PB}$

=> PA • PB = PC • PD