In the figure AB and CD are extended to intersect at P. If AB = 5, BP = 3, PD = 2 then find CD.

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Solution

$P A = 8, P B = 3, P D = 2$

$P A \times P B = P C \times P D$

$8 \times 3 = P C \times 2$

$P C = 12$
So,
$C D = P C - P D$
$= 12 - 2$
$= 10$

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