If two chords AB and CD intersect externally at P and if AB = 4, PB = 5, PD = 3, then find the value of CD.

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Solution

According BPT

We have,

$\frac{P C}{C D} = \frac{P B}{B A}$

$\frac{P C}{C D} = \frac{5}{4}$

Since PC = PD - CD

$\frac{P D - C D}{C D} = \frac{5}{4}$

$4 (P D - C D) = 5 C D$

$4 P D - 4 C D = 5 C D$

$4 P D = 5 C D + 4 C D$

$4 P D = 9 C D$

$4 (3) = 9 C D$ (Given PD = 3cm)

$12 = 9 C D$

$C D = \frac{12}{9}$

$C D = \frac{4}{3} = 1.3$

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