# Apollonius theorem

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

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## Apollonius' theorem

For a triangle \(ABC\) with \(M\) being the midpoint of \(\overline{BC}\), the following equation holds.

\[\large \overline{AB}^2+\overline{AC}^2=2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)^2\right\}\]

## Proof by Pythagorean Theorem

## Prove that a triangle \(ABC\) with \(M\) being the midpoint of \(\overline{BC}\) satisfies the apollonius theorem by only using pythagorean theorem.

Let \(H\) be a foot of perpendicular from \(A\) to \(\overline{BC}\).

It is clear that

\[\overline{BM}=\overline{CM}=\frac{\overline{BC}}{2} \\ \overline{BH}+\overline{CH}=\overline{BC}\]

Using Pythagorean Theorem, we get

\[\overline{AB}^2=\overline{AH}^2+\overline{BH}^2 \\ \overline{AC}^2=\overline{AH}^2+\overline{CH}^2 \\ \overline{AM}^2=\overline{AH}^2+\overline{MH}^2\]

From these, we can conclude that

\[\begin{align} \overline{AB}^2+\overline{AC}^2 &=2\overline{AH}^2+\overline{BH}^2+\overline{CH}^2 \\ &=2\overline{AH}^2+2\overline{MH}^2+\overline{BH}^2-\overline{MH}^2+\overline{CH}^2-\overline{MH}^2 \\ &=2\overline{AM}^2+\left(\overline{BH}+\overline{MH}\right)\left(\overline{BH}-\overline{MH}\right)+\left(\overline{CH}+\overline{MH}\right)\left(\overline{CH}-\overline{MH}\right) \\ &=2\overline{AM}^2+\left(\overline{BH}+\overline{MH}\right)\cdot\overline{BM}+\overline{CM}\cdot\left(\overline{CH}-\overline{MH}\right) \\ &=2\overline{AM}^2+\frac{\overline{BC}^2}{2} \\ &=2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)^2\right\} \end{align}\]

\(\square\)

## Proof by cosine rule

## Prove that a triangle \(ABC\) with \(M\) being the midpoint of \(\overline{BC}\) satisfies the apollonius theorem by only using cosine rule (law of cosines).

Using the cosine law, we get

\[\begin{align} \overline{AB}^2 & =\overline{BM}^2+\overline{AM}^2-2\overline{AM}\cdot\overline{BM}\cos\angle AMB \\ \overline{AC}^2 & =\overline{CM}^2+\overline{AM}^2-2\overline{AM}\cdot\overline{CM}\cos\angle AMC \\ & = \overline{BM}^2+\overline{AM}^2+2\overline{AM}\cdot\overline{BM}\cos\angle AMB~(\because \angle AMB+\angle AMC=\pi) \end{align}\]

Add them sides by sides.

\[\begin{align} \overline{AB}^2+\overline{AC}^2 & = 2\overline{AM}^2+2\cdot\left(\frac{\overline{BC}}{2}\right)^2 \\ & = 2\left\{\overline{AM}^2+\left(\frac{\overline{BC}}{2}\right)\right\} \end{align}\]

\(\square\)

## Proof by vectors

## Prove that a triangle \(ABC\) with \(M\) being the midpoint of \(\overline{BC}\) satisfies the apollonius theorem by only using elementary vector operations.

Let \(A\) be the origin of a Cartesian coordinates system.

Then define \(\overrightarrow{AB}=\vec{b}\) and \(\overrightarrow{AC}=\vec{c}.\)

It's clear that \(\overrightarrow{BC}=\vec{c}-\vec{b}\) and \(\overrightarrow{AM}=\dfrac{\vec{b}+\vec{c}}{2}.\)

Therefore,

\[\begin{align} \overline{AB}^2+\overline{AC}^2 &= |\vec{b}|^2+|\vec{c}|^2 \\ &= \dfrac{1}{2}(2|\vec{b}|^2+2|\vec{c}|^2) \\ &= \dfrac{1}{2}(|\vec{b}|^2+|\vec{c}|^2+2\vec{b}\cdot\vec{c}+|\vec{b}|^2+|\vec{c}|^2-2\vec{b}\cdot\vec{c}) \\ &= \dfrac{1}{2}\{(\vec{b}+\vec{c})^2+(\vec{c}-\vec{b})^2\} \\ &= \dfrac{1}{2}(4\overline{AM}^2+\overline{BC}^2) \\ &= 2\left\{\overline{AM}^2+\left(\dfrac{\overline{BC}}{2}\right)^2\right\}. \end{align}\]

\(\square\)

## Relations with other theorems

Apollonius' theoremis a special case of stewart's theorem, and also a generalization of pythagorean theorem.

Substitute \(\overline{BP}=\overline{CP}\) to 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, if \(\angle BAC=\dfrac{\pi}{2}\), \(\overline{AM}=\dfrac{\overline{BC}}{2}\). Substitute that to Apollonius' Theorem, and you get

\[\begin{align} \overline{AB}^2+\overline{AC}^2 & =2\left\{\left(\dfrac{\overline{BC}}{2}\right)^2+\left(\dfrac{\overline{BC}}{2}\right)^2\right\} \\ & = \overline{BC}^2 \end{align}\]

\(\square\)

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 centroid of a triangle for proof.

## Examples

Find \( a + b + d + 1 \).

**Cite as:**Apollonius theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/apollonius-theorem/