The symmedian and the tangents at B and C

Setup. $\angle A$ acute $\Rightarrow$ the tangents at $B$ and $C$ meet at $T$ on the other side of $BC$ from $A$; $TB=TC$. By the tangent–chord theorem, $\angle TBC=\angle TCB=\angle A$.

Angles. $\angle ABT=\angle ABC+\angle CBT=B+A$, so $\sin\angle ABT=\sin C$; similarly $\sin\angle ACT=\sin B$.

Area ratio. $D=AT\cap BC$ so
$$\frac{BD}{DC}=\frac{[ABT]}{[ACT]}=\frac{c\cdot\sin C}{b\cdot\sin B}=\frac{c^2}{b^2},$$
using the law of sines.

Why "symmedian". The median gives ratio $1$, the bisector $\frac cb$, the symmedian $\frac{c^2}{b^2}$: the reflection of the median in the bisector.

Remark. If $\angle A$ is obtuse, the same works. For $\angle A=90^\circ$ the tangents are parallel and $T$ does not exist.

Lesson. A ratio of segments on a line equals a ratio of areas with a common base; the tangent–chord angle then makes the sines match the sides.