In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) \(\triangle\) APC ~ \(\triangle\) DPB
(ii) AP . PB = CP . DP

Firstly, let us join CB, in the given figure.

(i) In \(\triangle\) APC and \(\triangle\) DPB,
\(\angle\) APC = \(\angle\) DPB (Vertically opposite angles)
\(\angle\) CAP = \(\angle\) BDP (Angles in the same segment for chord CB)

Therefore,
\(\triangle\) APC ~ \(\triangle\) DPB (AA similarity criterion)


(ii) In the above, we have proved that \(\triangle\) APC ~ \(\triangle\) DPB

We know that the corresponding sides of similar triangles are proportional.
\(\frac{AP}{DP} = \frac{PC}{PB} = \frac{CA}{BD} \)
\(\frac{AP}{DP} = \frac{PC}{PB} \)
AP. PB = PC. DP
Hence, proved.