Apollonius's Theorem

Apollonius's theorem is an elementary geometry theorem relating the length of a median of a triangle to the lengths of its sides. While most of the world refers to it as it is, in East Asia, the theorem is usually referred to as Pappus's theorem or midpoint theorem. It can be proved by Pythagorean theorem from the cosine rule as well as by vectors. The theorem is named after a Greek mathematician Apollonius of Perga.

For triangle $ABC$ with $M$ being the midpoint of $\overline{BC}$, the following equation holds:

$\overline{AB}^2+\overline{AC}^2=2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)^2\right\}.$

Prove that triangle $ABC,$ with $M$ being the midpoint of $\overline{BC},$ satisfies Apollonius' theorem by using only the Pythagorean theorem.

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Let $H$ be the foot of perpendicular from $A$ to $\overline{BC}$.

It is clear that

$\begin{aligned} \overline{BM}=\overline{CM}&=\frac{\overline{BC}}{2} \\\\ \overline{BH}+\overline{CH}&=\overline{BC}. \end{aligned}$

Using the Pythagorean theorem, we get

$\begin{aligned} \overline{AB}^2&=\overline{AH}^2+\overline{BH}^2 \\ \overline{AC}^2&=\overline{AH}^2+\overline{CH}^2 \\ \overline{AM}^2&=\overline{AH}^2+\overline{MH}^2. \end{aligned}$

From these, we can conclude that

$\begin{aligned} \overline{AB}^2+\overline{AC}^2 &=2\overline{AH}^2+\overline{BH}^2+\overline{CH}^2 \\ &=2\overline{AH}^2+2\overline{MH}^2+\overline{BH}^2-\overline{MH}^2+\overline{CH}^2-\overline{MH}^2 \\ &=2\overline{AM}^2+\left(\overline{BH}+\overline{MH}\right)\left(\overline{BH}-\overline{MH}\right)+\left(\overline{CH}+\overline{MH}\right)\left(\overline{CH}-\overline{MH}\right) \\ &=2\overline{AM}^2+\left(\overline{BH}+\overline{MH}\right)\cdot\overline{BM}+\overline{CM}\cdot\left(\overline{CH}-\overline{MH}\right) \\ &=2\overline{AM}^2+\frac{\overline{BC}^2}{2} \\ &=2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)^2\right\}.\ _\square \end{aligned}$

Prove that triangle $ABC,$ with $M$ being the midpoint of $\overline{BC},$ satisfies Apollonius's theorem by using only the cosine rule (law of cosines).

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Using the cosine rule, we get

$\begin{aligned} \overline{AB}^2 & =\overline{BM}^2+\overline{AM}^2-2\overline{AM}\cdot\overline{BM}\cos\angle AMB \\\\ \overline{AC}^2 & =\overline{CM}^2+\overline{AM}^2-2\overline{AM}\cdot\overline{CM}\cos\angle AMC \\ & = \overline{BM}^2+\overline{AM}^2+2\overline{AM}\cdot\overline{BM}\cos\angle AMB. \qquad(\text{since } \angle AMB+\angle AMC=\pi) \end{aligned}$

Adding these two gives

$\begin{aligned} \overline{AB}^2+\overline{AC}^2 & = 2\overline{AM}^2+2\overline{BM}^2 \\ & = 2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)^2\right\}.\ _\square \end{aligned}$

Prove that triangle $ABC,$ with $M$ being the midpoint of $\overline{BC},$ satisfies Apollonius's theorem only by using elementary vector operations.

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Let $A$ be the origin of a Cartesian coordinate system.

Defining $\overrightarrow{AB}=\vec{b}$ and $\overrightarrow{AC}=\vec{c},$ it's clear that $\overrightarrow{BC}=\vec{c}-\vec{b}$ and $\overrightarrow{AM}=\frac{\vec{b}+\vec{c}}{2}.$

Therefore,

$\begin{aligned} \overline{AB}^2+\overline{AC}^2 &= |\vec{b}|^2+|\vec{c}|^2 \\ &= \dfrac{1}{2}\big(2|\vec{b}|^2+2|\vec{c}|^2\big) \\ &= \dfrac{1}{2}\big(|\vec{b}|^2+|\vec{c}|^2+2\vec{b}\cdot\vec{c}+|\vec{b}|^2+|\vec{c}|^2-2\vec{b}\cdot\vec{c}\big) \\ &= \dfrac{1}{2}\left\{\big(\vec{b}+\vec{c})^2+(\vec{c}-\vec{b}\big)^2\right\} \\ &= \dfrac{1}{2}\left(4\overline{AM}^2+\overline{BC}^2\right) \\ &= 2\left\{\overline{AM}^2+\left(\dfrac{\overline{BC}}{2}\right)^2\right\}.\ _\square \end{aligned}$

Apollonius's theorem is a special case of Stewart's theorem and also a generalization of the Pythagorean theorem.

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Substitute $\overline{BP}=\overline{CP}$ into Stewart's theorem, $\overline{CP}\cdot\overline{AB}^2+\overline{BP}\cdot\overline{AC}^2=\left(\overline{BP}+\overline{CP}\right)\left(\overline{AP}^2+\overline{BP}\cdot\overline{CP}\right)$, and you get

$\overline{AB}^2+\overline{AC}^2=2\left(\overline{AP}^2+\overline{BP}^2\right).$

Also, substituting $\angle BAC=\frac{\pi}{2}$ and $\overline{AM}=\frac{\overline{BC}}{2}$ into Apollonius's theorem, you get

$\begin{aligned} \overline{AB}^2+\overline{AC}^2 & =2\left\{\left(\frac{\overline{BC}}{2}\right)^2+\left(\frac{\overline{BC}}{2}\right)^2\right\} \\ & = \overline{BC}^2.\ _\square \end{aligned}$

We can also obtain the Carnot's theorem, which reads

$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2$

for any point $P.$ Look for the Centroid of a Triangle wiki for proof.