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.

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

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

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

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

\(\square\)

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.

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

\(\square\)

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,

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

\(\square\)

Apollonius' theorem is 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.