OlympiadMediumGeometry10–12

Ptolemy on an Equilateral Triangle

Ptolemeu num Triângulo Equilátero

An equilateral triangle $ABC$ is inscribed in a circle, and $P$ is a point on the arc $BC$ that does not contain $A$ . Prove that $PA = PB + PC$ .

The four points lie on one circle; recall the identity Ptolemy gives for a cyclic quadrilateral.

Solution

Step 1 of 8

1.The points $A$ , $B$ , $P$ , $C$ lie on the circle in that order, so $ABPC$ is a cyclic quadrilateral with diagonals $AP$ and $BC$ .