# In the 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

## Solution :

(i)  In $$\triangle$$s PAC and PDB, we have :

$$\angle$$ APC = $$\angle$$ DPB         (vertically opp. angles)

$$\angle$$ CAP = $$\angle$$ BDP        (angles in same segment of circle are equal)

$$\therefore$$  By AA similarity, we have :

$$\triangle$$ APC ~ $$\triangle$$ DPB

(ii)  Since $$\triangle$$s APC ~ DPB, therefore

$$AP\over DP$$ = $$CP\over PB$$  or  PA.PB = CP.DP.