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.


1 Answer 1


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 $\triangle AMB'$).

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By the Law of Sines in $\triangle AMC$ and $\triangle AMB'$, we have $$\frac{\sin\gamma}{\sin(\beta+\theta)}\cdot\frac{\sin(\beta-\theta)}{\sin \beta} \;=\;\frac{d}{b}\cdot\frac{b\cos\alpha}{d} \;=\; \cos\alpha$$


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