# 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.

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## Apollonius's Theorem

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

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

## Proof by Pythagorean Theorem

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

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

It is clear that

\[\begin{align} \overline{BM}=\overline{CM}&=\frac{\overline{BC}}{2} \\\\ \overline{BH}+\overline{CH}&=\overline{BC}. \end{align}\]

Using the Pythagorean theorem, we get

\[\begin{align} \overline{AB}^2&=\overline{AH}^2+\overline{BH}^2 \\ \overline{AC}^2&=\overline{AH}^2+\overline{CH}^2 \\ \overline{AM}^2&=\overline{AH}^2+\overline{MH}^2. \end{align}\]

From these, we can conclude that

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

## Proof by Cosine Rule

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).

Using the cosine rule, we get

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

Adding these two gives

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

## Proof by Vectors

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

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,

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

## Relations with Other Theorems

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

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

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

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.

## Examples

Triangle \(ABC\) has side lengths \(\overline{AB}=2\sqrt{3}\) and \(\overline{BC}=2,\) as shown. \(D\) is the midpoint of \(\overline{BC},\) satisfying \(\overline{AD}=\sqrt{7}.\)

Let \(E\) be the point of intersection between \(\overline{AB}\) and the bisector of \(\angle ACB.\) \(\overline{CE}\) meets with \(\overline{AD}\) at point \(P,\) and the bisector of \(\angle APE\) meets with \(\overline{AB}\) at point \(R.\) An extension of \(\overline{PR}\) meets with \(\overline{BC}\) at point \(Q.\)

Given that the area of \(\triangle PQC\) is \(a+b\sqrt{7}\) times larger than that of \(\triangle PRE\) for some rational numbers \(a\) and \(b,\) find the value of \(ab.\)

*This problem is a part of <Grade 10 CSAT Mock Test> series.*

Triangle \(ABC\) with its centroid at \(G\) has side lengths \(AB=15, BC=18,AC=25\). \(D\) is the midpoint of \(BC\).

The length of \(GD\) can be expressed as \( \frac{ a \sqrt{d} } { b} \), where \(a\) and \(b\) are coprime positive integers and \(d\) is a square-free positive integer.

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

**Cite as:**Apollonius's Theorem.

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