# Stewart's Theorem

In geometry, **Stewart's theorem** yields a relation between the side lengths and a cevian length of a triangle. It can be proved from the *law of cosines* as well as by the famous *Pythagorean theorem*. Its name is in honor of the Scottish mathematician Matthew Stewart who published the theorem in 1746 when he was believed to be a candidate to replace Colin Maclaurin as Professor of Mathematics at the University of Edinburgh.

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## Stewart's Theorem

In $\triangle ABC$, point $D$ is a point on $BC$ and $AB=c, AC=b, BD=u, DC=v, AD=t.$ Stewart's theorem states that in this triangle, the following equation holds:

$t^2=\frac{b^2u+c^2v}{u+v}-uv.$

## Proof by the Law of Cosines

By the law of cosines, we have $\begin{aligned} b^2&=v^2+t^2-2vt\cos \theta &\qquad (1)\\ c^2&=u^2+t^2+2ut\cos \theta. &\qquad (2) \end{aligned}$ Now multiply (1) by $u$ and multiply (2) by $v$ to eliminate $\cos \theta$: $\begin{aligned} b^2u&=uv^2+ut^2-2uvt\cos \theta &\qquad (3)\\ c^2v&=u^2v+vt^2+2uvt\cos \theta. &\qquad (4) \end{aligned}$ Taking $(3)+(4)$ gives $\begin{aligned} b^2u+c^2v&=uv(u+v)+t^2(u+v)\\ \Rightarrow t^2&=\frac{b^2u+c^2v}{u+v}-uv.\ _\square \end{aligned}$

Stewart's theorem can sometimes be rewritten as $b^2u+c^2v=(u+v)(uv+t^2)$.

## Proof by the Pythagorean Theorem

The proof below assumes $\angle B$ and $\angle C$ are both acute and $u < v$ as in the figure above. Then we have $\begin{aligned} t^2 &= h^2 + x^2 \\ b^2 &= h^2 + (v-x)^2 \Rightarrow b^2u = h^2u + uv^2 - 2uvx + ux^2 \\ c^2 &= h^2 + (u+x)^2 \Rightarrow c^2v = h^2v + u^2v + 2uvx + vx^2, \end{aligned}$ which implies $\begin{aligned} b^2u + c^2v &= h^2u + h^2v + uv^2 + u^2v - 2uvx + 2uvx + ux^2 + vx^2 \\ &= (u + v)(h^2 + uv + x^2) \\ &= (u + v)(t^2 + uv) \\ &= a \cdot (t^2 + uv). \ _\square \end{aligned}$

## Special case where $\Delta ABC$ is Isosceles

In the case where $\Delta ABC$ is isosceles (see figure above), Stewart's theorem has a more simplified form: $\begin{aligned} a \cdot (t^2 + uv) &= b^2u + c^2v \\ &= b^2u + b^2v \\ &= b^2 (u + v) \\ &= ab^2 \\ \Rightarrow b^2 &= t^2 + uv. \end{aligned}$ This theorem is quite useful in calculating the length of standard cevians like median, angle bisector, etc.

## Additional Problems

Let $ABCD$ be a square, and let $E$ and $F$ be points on $AB$ and $BC,$ respectively. The line through $E$ parallel to $BC$ and the line through $F$ parallel to $AB$ divide $ABCD$ into two squares and two non-square rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$

Find $\frac{AE}{EB}+\frac{EB}{AE}.$

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

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