<s>[INST] Point inside a right angled triangle A right angled triangle $ABC$ $(\measuredangle C=90^\circ)$ is given with $\measuredangle BAC=\alpha$. The point $O$ lies inside the triangle $ABC$ such that $\measuredangle OAB=\measuredangle OBC=\measuredangle OCA=\varphi$. Show that $\tan\varphi=\sin\alpha\cdot\cos\alpha.$
We can write the RHS of the equality that we are supposed to prove as $$\sin\alpha\cdot\cos\alpha=\dfrac{a}{c}\cdot\dfrac{b}{c}=\dfrac{ab}{c^2}$$
So we can try to show the equality $$\tan\varphi=\dfrac{ab}{c^2}$$
To express $\tan\varphi$ in some way, all I can think of is to include $\measuredangle\varphi$ in a right triangle and then use the definition of tangent of an acute angle. For this purpose, let's draw perps from $O$ to the sides of the triangle $ABC$. Their foots are $H, H_1, H_2$ on $AB,BC$ and $AC$, respectively. Then we have $$\tan\varphi=\dfrac{OH}{AH}=\dfrac{OH_1}{BH_1}=\dfrac{OH_2}{CH_2}$$
This seems pointless. Thank you in advance!
 [/INST] Let the hypotenuse $AB = 1$. Then $BC=\sin\alpha$.
Consider $\triangle OAB$. $\angle OAB = \varphi$ and $\angle ABO = (90^\circ-\alpha) - \varphi$, so $\angle BOA = 90^\circ + \alpha$. By the laws of sine,
$$\begin{align*}
\frac{OB}{\sin \angle OAB} &= \frac{AB}{\sin\angle BOA}\\
OB &= \frac{1\cdot\sin \varphi}{\sin (90^\circ + \alpha)}\\
&= \frac{\sin \varphi}{\cos \alpha}\\
\end{align*}$$
Then consider $\triangle OBC$. $\angle OBC = \varphi$ and $\angle BCO = 90^\circ - \varphi$, so $\angle COB = 90^\circ$.
$$\begin{align*}
\cos \angle OBC = \cos\varphi &= \frac{OB}{BC}\\
\cos\varphi &= \frac{\sin\varphi}{\cos \alpha} \cdot \frac{1}{\sin\alpha}\\
\sin\alpha\cos\alpha &= \frac{\sin\varphi}{\cos \varphi}\\
\tan \varphi&= \sin\alpha \cos \alpha
\end{align*}$$
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