Exercise 1260Geometry › Plane geometryHigh schoolAdded 4 Nov 2026

Ptolemy and a point on the circumcircle of an equilateral triangle

#Ptolemy's theorem#cyclic quadrilateral#equilateral triangle

Statement

Let $ABC$ be an equilateral triangle and let $P$ be a point on the arc $BC$ of its circumcircle that does not contain $A$ . Prove that
$PA=PB+PC.$
Deduce that $PB+PC\le 2R$ , where $R$ is the circumradius, and find when equality holds.

Hint

You may use Ptolemy's theorem: in a cyclic quadrilateral $WXYZ$ (vertices in this order on the circle), $WY\cdot XZ=WX\cdot YZ+XY\cdot ZW$ . Apply it to the quadrilateral $ABPC$ .

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Solution

Idea. The quadrilateral $ABPC$ has three sides equal to the side $s$ of the triangle, and Ptolemy turns this into a linear relation.

Applying Ptolemy. The points $A,B,P,C$ lie on the circle in this order. Ptolemy gives
$AP\cdot BC=AB\cdot PC+AC\cdot BP.$
Since $AB=BC=CA=s$ , we get $s\cdot PA=s\cdot PC+s\cdot PB$ , so $PA=PB+PC$ . $\blacksquare$

The inequality. $PA$ is a chord of the circle, so $PA\le 2R$ . Hence $PB+PC=PA\le2R$ . Equality holds iff $PA$ is a diameter, i.e. $P$ is the point diametrically opposite to $A$ on the arc $BC$ ; then $PB=PC=R$ .

Remark. The identity $PA=PB+PC$ also shows that the three segments $PA,PB,PC$ always form a (degenerate) triangle.

Lesson. Look for Ptolemy whenever a cyclic quadrilateral has many equal sides; it converts products of lengths into linear relations.

Ptolemy and a point on the circumcircle of an equilateral triangle

Statement

Solution

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