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Power of a point with respect to a circle

Submitted by Structure on Sat, 11/29/2008 - 18:29.

Two lines by a point P outside a circle cut in A,B respectively in C and D. Then

$$PA*PB=PC*PD=TP^2$$
$$\triangle PAC\sim\triangle PDB$$

as $ \angle PBD=\angle PCA\:\:;\angle PDB=\angle PAC $

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

Average: 2.3 (4 votes)
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