The median-length theorem and computing a median

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Let $ABC$ be a triangle and let $AM$ be the median to side $BC$ ($M$ the midpoint of $BC$).

• (a) Prove the median-length theorem: the sum of the squares of two sides equals twice the sum of the square of half the third side and the square of the median to it, that is $\overline{AB}^{\,2}+\overline{AC}^{\,2}=2\left[\left(\frac{\overline{BC}}{2}\right)^2+\overline{AM}^{\,2}\right].$
• (b) Given $\overline{BC}=9$, $\overline{CA}=8$, $\overline{AB}=7$, compute the length of the median $\overline{AM}$.

Solution

(a) Proof. Set $\overline{AM}=m$, $\overline{BM}=\overline{MC}=\dfrac{\overline{BC}}{2}$ and let $\theta=\widehat{AMB}$; then $\widehat{AMC}=180^\circ-\theta$. Apply the law of cosines to the two triangles $ABM$ and $ACM$: $\overline{AB}^{\,2}=m^2+\left(\frac{\overline{BC}}{2}\right)^2-2\,m\,\frac{\overline{BC}}{2}\cos\theta,$ $\overline{AC}^{\,2}=m^2+\left(\frac{\overline{BC}}{2}\right)^2-2\,m\,\frac{\overline{BC}}{2}\cos(180^\circ-\theta).$ Since $\cos(180^\circ-\theta)=-\cos\theta$, on adding the two equations the cosine terms cancel: $\overline{AB}^{\,2}+\overline{AC}^{\,2}=2m^2+2\left(\frac{\overline{BC}}{2}\right)^2=2\left[\left(\frac{\overline{BC}}{2}\right)^2+\overline{AM}^{\,2}\right].$ This is the median-length theorem. $\;\blacksquare$

(b) Numerical computation. Solve for $\overline{AM}$: $\overline{AM}^{\,2}=\frac{\overline{AB}^{\,2}+\overline{AC}^{\,2}}{2}-\left(\frac{\overline{BC}}{2}\right)^2=\frac{7^2+8^2}{2}-\left(\frac{9}{2}\right)^2=\frac{49+64}{2}-\frac{81}{4}.$ $\overline{AM}^{\,2}=\frac{113}{2}-\frac{81}{4}=\frac{226-81}{4}=\frac{145}{4}=36{,}25.$ $\overline{AM}=\sqrt{36{,}25}=\frac{\sqrt{145}}{2}\approx 6{,}021.$ $\boxed{\overline{AM}=\frac{\sqrt{145}}{2}\approx 6{,}021}$