Stewart's theorem and its consequences

1. Let $\theta=\angle ADB$, so $\angle ADC=180^\circ-\theta$ and $\cos\angle ADC=-\cos\theta$. The law of cosines gives
$$c^2=d^2+m^2-2dm\cos\theta,\qquad b^2=d^2+n^2+2dn\cos\theta.$$
Multiply the first by $n$, the second by $m$, and add: the cosine terms cancel and
$$c^2n+b^2m=d^2(m+n)+mn(m+n)=a\,(d^2+mn).$$

2. For the median, $m=n=\frac a2$: $\frac a2(b^2+c^2)=a\left(d^2+\frac{a^2}4\right)$, hence $d^2=\frac{b^2+c^2}2-\frac{a^2}4=\frac{2b^2+2c^2-a^2}4$.

3. By the angle bisector theorem $\frac mn=\frac cb$, so $m=\frac{ac}{b+c}$ and $n=\frac{ab}{b+c}$. Then
$$b^2m+c^2n=\frac{abc\,(b+c)}{b+c}=abc,$$
and Stewart gives $d^2=bc-mn=bc-\frac{a^2bc}{(b+c)^2}=bc\left(1-\frac{a^2}{(b+c)^2}\right)$.

4. Median: $\frac{2\cdot225+2\cdot169-196}4=148$, so the median is $\sqrt{148}=2\sqrt{37}\approx12.17$. Bisector: $\ell^2=195\left(1-\frac{196}{784}\right)=195\cdot\frac34=146.25$, so $\ell\approx12.09$.

Remark. Stewart's theorem also gives the length of an altitude or any cevian once $m$ and $n$ are known.

Lesson. When two angles are supplementary, write the law of cosines in both triangles and eliminate the cosine.