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Cyclic Quadrilateral and its Property

A line drawn ...

Question

A line drawn from the vertex A of an equilateral triangle ABC meets BC at D and the circumcircle at P.

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Solution

$Δ B A C \to$ equilateral

$A P B C \to$ Cyclic Quadrilateral
$A B = A C = B C = a$
Using Ptolemy’s Theorem in cyclic quadrilateral $A B C P$ :
$P A . B C + B P . A C + P C . A B$
$P A . a/ + P B . a/ + P C . a/$

Hence,
$P A = P B + P C$ .
Option C is correct.

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