# Proving a trig identity with a median in triangle ABC

In a given triangle $$ABC$$, $$\angle BAC=\alpha, \angle ABC = \beta, \angle ACB = \gamma, \angle BAM = \theta$$, where $$M$$ is the midpoint of side $$BC$$.

Prove the identity $$\frac{\sin\gamma}{\sin\beta}\cdot\frac{\sin\left(\beta-\theta\right)}{\sin\left(\beta+\theta\right)}=\cos\alpha.$$

 I have checked this identity empirically, but I am having trouble proving the general statement. I tried to get an expression for $$\sin\theta$$ by realizing that  $$2 \cdot \frac{1}{2} AB \cdot AM \cdot \sin\theta = \frac{1}{2} AB \cdot AC \cdot \sin\alpha$$  since the area of $$\triangle ABC$$ is twice the area of $$\triangle ABM$$. However, this leads to an ungodly mess when computing $$AM$$ using Stewart's theorem and using the sine compound angle formula on the LHS fraction.  My question: is there a clean(ish) way to prove this identity without resorting to extremely messy computations?

My motivation: the proof of this identity leads to a proof of a symmedian property I am investigating.

For non-obtuse $$\alpha$$ (the obtuse case is comparable), we can argue as follows:
• By the Exterior Angle Theorem in $$\triangle AMB$$, we have $$\angle AMC=\beta+\theta$$.
• Drop a perpendicular from $$C$$ to $$B'$$ on $$\overline{AB}$$, so that $$|AB'|=b\cos\alpha$$. Note that $$|MB'|=|MB|=|MC|$$, whence $$\angle BB'M=\beta$$, and thus $$\angle AMB'=\beta-\theta$$ (by the Exterior Angle Theorem in $$\triangle AMB'$$).
By the Law of Sines in $$\triangle AMC$$ and $$\triangle AMB'$$, we have $$\frac{\sin\gamma}{\sin(\beta+\theta)}\cdot\frac{\sin(\beta-\theta)}{\sin \beta} \;=\;\frac{d}{b}\cdot\frac{b\cos\alpha}{d} \;=\; \cos\alpha$$