# Stewart's Theorem

## Theorem

Let $\triangle ABC$ be a triangle with sides $a, b, c$.

Let $CP$ be a cevian from $C$ to $P$. Then:

$a^2 \cdot AP + b^2 \cdot PB = c \paren {CP^2 + AP \cdot PB}$

## Proof

 $\text {(1)}: \quad$ $\ds b^2$ $=$ $\ds AP^2 + CP^2 - 2 AP \cdot CP \cdot \map \cos {\angle APC}$ Law of Cosines $\text {(2)}: \quad$ $\ds a^2$ $=$ $\ds PB^2 + CP^2 - 2 CP \cdot PB \cdot \map \cos {\angle BPC}$ Law of Cosines $\ds$ $=$ $\ds PB^2 + CP^2 + 2 CP \cdot PB \cdot \map \cos {\angle APC}$ Cosine of Supplementary Angle $\text {(3)}: \quad$ $\ds \leadsto \quad \ \$ $\ds b^2 \cdot PB$ $=$ $\ds AP^2 \cdot PB + CP^2 \cdot PB - 2 PB \cdot AP \cdot CP \cdot \map \cos {\angle APC}$ $(1) \ \times PB$ $\text {(4)}: \quad$ $\ds a^2 \cdot AP$ $=$ $\ds PB^2 \cdot AP + CP^2 \cdot AP + 2 AP \cdot CP \cdot PB \cdot \map \cos {\angle APC}$ $(2) \ \times AP$ $\ds \leadsto \quad \ \$ $\ds a^2 \cdot AP + b^2 \cdot PB$ $=$ $\ds AP^2 \cdot PB + PB^2 \cdot AP + CP^2 \cdot PB + CP^2 \cdot AP$ $(3) \ + \ (4)$ $\ds$ $=$ $\ds CP^2 \left({PB + AP}\right) + AP \cdot PB \paren {PB + AP}$ $\ds$ $=$ $\ds c \paren {CP^2 + AP \cdot PB}$ as $PB + AP = c$

$\blacksquare$

## Source of Name

This entry was named for Matthew Stewart.

It is also known as Apollonius's Theorem after Apollonius of Perga.