, 2014

Let $M$, $N$ be the feet of the perpendiculars from $A$, $D$ on $BC$. Then, $N$ is the reflection of $M$ and the right triangles $AMB$ and $DNC$ are congruent. Moreover, $AD$ is parallel to $BC$ and $|AD| = |MN|$. There are various possibilities for the position of $M$, some are shown below.

We let $a = |BC|$, $b = |CA|$ and $c = |AB|$ as usual. Let $x$ denote the signed distance from $M$ to the midpoint $E$ of $BC$. The sign is chosen such that $x = -|ME| \le 0$ if $E$ lies between $B$ and $M$, and $x = |ME| \ge 0$ otherwise.

For all possible positions of $M$ on the line $BC$ we then have
$$|MN| = 2|x|, \quad |CM| = \left|\frac{a}{2} + x\right| \quad \text{and} \quad |BM| = \left|\frac{a}{2} - x\right|.$$
The Theorem of Pythagoras gives
$$b^2 = |AM|^2 + |CM|^2 \quad \text{and} \quad c^2 = |AM|^2 + |BM|^2.$$
The Cosine Theorem $b^2 = a^2 + c^2 - 2ac \cos \angle ABC$ implies now
$$a(a - 2c \cos \angle ABC) = b^2 - c^2 = \left(\frac{a}{2} + x\right)^2 - \left(\frac{a}{2} - x\right)^2 = 2ax,$$
hence $|AD| = |MN| = 2|x| = |a - 2c \cos \angle ABC|$, as required.