Stewart's Theorem

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Theorem

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

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

Stewart's Theorem.png

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.