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

## Solution :

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

$$\angle$$ APC = $$\angle$$ DPB         (common)

$$\angle$$ PAC = $$\angle$$ PDB

[$$\therefore$$  $$\angle$$ BAC = 180 – $$\angle$$ PAC  and $$\angle$$ PDB = $$\angle$$ CDB = 180 – (180 – $$\angle$$ PAC) = $$\angle$$ PAC]

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

$$\triangle$$ PAC ~ $$\triangle$$ PDB

(ii)  Since $$\triangle$$s PAC ~ PDB, therefore

$$PA\over PD$$ = $$PC\over PB$$  or  PA.PB = PC.PD.