# Quadrilateral ABCD

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• March 13th 2010, 02:53 PM
disclaimer
Quadrilateral ABCD
Quadrilateral $ABCD$ is inscribed in a circle, where the tangents to this circle at points $B$ and $D$ intersect on the straight line joining $A$ and $C$. Show that:

$AB\cdot{CD}=AD\cdot{BC}$

Thanks in advance.
• March 14th 2010, 10:34 PM
sa-ri-ga-ma
Quote:

Originally Posted by disclaimer
Quadrilateral $ABCD$ is inscribed in a circle, where the tangents to this circle at points $B$ and $D$ intersect on the straight line joining $A$ and $C$. Show that:

$AB\cdot{CD}=AD\cdot{BC}$

Thanks in advance.

If P is the point of intersection of two tangents at B and D and the line CA, then
PB = PD. You can easily show that the triangles PBA and PBD are congruent. Similarly Triangles PBC and PDC arfe also congruent.
Now proceed further to get the required result.