Cevian Sine-Ratio

Cevian sine-ratio (Ratio Lemma)

Consider a triangle with a cevian drawn from vertex $A$ to side $BC$. For example, consider the following diagram.

We want to learn more about the point $D$! Specifically, we would like to know how point $D$ relates with respect to the side $BC$. This ratio will end up coming in handy in several classical proofs and problems.

How would we find the ratio $BD/DC$? Think about this a bit before continuing. There are a couple ways to proceed.

If you have an idea for a proof by now, great!

If not, here's a leading question towards a natural way to compute $\tfrac{BD}{DC}$. What else does the ratio represent?

$\tfrac{BD}{DC}$ is the ratio of the areas $[ABD]$ and $[ACD]$! This is because both triangles have the same height. We can express the areas of the triangles with the sine area formula: $$[ABD] = \frac12 AB\cdot AD\sin \angle BAD$$ $$[ACD] = \frac12 AC \cdot AD \sin \angle DAC$$ When we divide the two areas, we can see that $\tfrac12 AD$ terms cancel out: $$\frac{[ABD]}{[ACD]} = \frac{\tfrac12 AB \cdot AD \sin\angle BAD}{\tfrac12 AC \cdot AD \sin \angle DAC} = \frac{AB}{AC} \cdot \frac{\sin \angle BAD}{\sin \angle DAC} = \frac{BD}{DC}$$ So we have a nice expression for the ratio $\frac{BD}{DC}$ at this point. This ratio, sometimes called the Ratio Lemma, will come in handy in several proofs, notably in the trigonometric form of Trig Ceva.

Notice that we can also prove the ratio with the Law of Sines on triangles $ABD$ and $ACD$--either proof works. If you were thinking of this proof, great!