Question

In Fig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) $△PAC\sim △PDB$
(ii) $PA.PB=PC.PD$

Answer 1

(i) In $△PAC$ and $△PDB$,

$\angle BAC={180}^{\circ }-\angle PAC$ (linear pairs)

$\angle PDB=\angle CDB={180}^{\circ }-\angle BAC$

$={180}^{\circ }-({180}^{\circ }-\angle PAC)=\angle PAC$)

$\angle PAC=\angle PDB$ ... (1)

$\angle APC=\angle BPD$ [Common] ... (2)

We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
Therefore, for cyclic quadrilateral ABCD.
consider ABCD $\Delta PAC$ and $\Delta PBD.$
$\angle PCA=\angle PBD$ .... (3)

$\therefore$ By AAA-criterion of similarity,

$△PAC\sim △DPB$

(ii) $△PAC\sim △DPB$

So, the corresponding sides of the similar triangles are proportional

$\frac{PA}{PD}=\frac{PC}{PB}$

$\Rightarrow PA.PB=PC.PD$