# 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 \(BC\), \(AB=c\), \(AC=b\), \(BD=u\), \(DC=v\), \(AD=t\), 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 \[\begin{align} b^2&=v^2+t^2-2vt\cos \theta &\qquad (1)\\ c^2&=u^2+t^2+2ut\cos \theta. &\qquad (2) \end{align}\] Now multiply (1) by \(u\) and multiply (2) by \(v\) to eliminate \(\cos \theta\): \[\begin{align} b^2u&=uv^2+ut^2-2uvt\cos \theta &\qquad (3)\\ c^2v&=u^2v+vt^2+2uvt\cos \theta. &\qquad (4) \end{align}\] Taking \((3)+(4)\) gives \[\begin{align} b^2u+c^2v&=uv(u+v)+t^2(u+v)\\ \Rightarrow t^2&=\frac{b^2u+c^2v}{u+v}-uv.\ _\square \end{align}\]

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 \[\begin{align} t^2 &= h^2 + x^2 \\ b^2 &= h^2 + (v-x)^2 \Rightarrow b^2u = h^2u + uv^2 - 2uvx + ux^2 \\ c^2 &= h^2 + (u+x)^2 \Rightarrow c^2v = h^2v + u^2v + 2uvx + vx^2, \end{align} \] which implies \[\begin{align} b^2u + c^2v &= h^2u + h^2v + uv^2 + u^2v - 2uvx + 2uvx + ux^2 + vx^2 \\ &= (u + v)(h^2 + uv + x^2) \\ &= (u + v)(t^2 + uv) \\ &= a \cdot (t^2 + uv). \ _\square \end{align} \]

## 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: \[\begin{align} a \cdot (t^2 + uv) &= b^2u + c^2v \\ &= b^2u + b^2v \\ &= b^2 (u + v) \\ &= ab^2 \\ \Rightarrow b^2 &= t^2 + uv. \end{align} \] This theorem is quite useful in calculating the length of standard cevians like median, angle bisector, etc.

## Additional Problems

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

In \(\triangle ABC\), \(AC=BC\), point \(D\) is on \(BC\) such that \(CD=3\times BD\), \(E\) is the midpoint of \(AD\) and that \(CE=\sqrt 7\) and \(BE=3\).

If the area of \(\triangle ABC\) is \(m\sqrt n\), where \(m\) and \(n\) are positive integers and \(n\) is square-free. Find \(m+n\).

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 \(\dfrac{9}{10}\) of the area of the square \(ABCD.\) Find \(\dfrac{AE}{EB}+\dfrac{EB}{AE}.\)

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

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