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!
 A: 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 $\angle COB = 90^\circ$.
$$\begin{align*}
\cos \angle OBC = \cos\varphi &= \frac{OB}{BC}\\
\cos\varphi &= \frac{\sin\varphi}{\cos \alpha} \cdot \frac{1}{\sin\alpha}\\
\sin\alpha\cos\alpha &= \frac{\sin\varphi}{\cos \varphi}\\
\tan \varphi&= \sin\alpha \cos \alpha
\end{align*}$$
A: 
Since $\angle OCB=90^\circ-\phi$, we have that $\angle BOC$ is a right angle, so that by Thales' Theorem, $O$ lies on a semicircle with diameter $\overline{BC}$. Extend $\overline{AO}$ to meet the other semicircle at $A'$, and we have $\angle BA'C$ is a right angle. Also, since $\angle OBC$ and $\angle OA'C$ are inscribed angles subtending the common arc $\stackrel{\frown}{OC}$, they are congruent. This in turn makes $\overline{AB}\parallel\overline{A'C}$, so that $\overline{A'B}$ is perpendicular to both lines; in particular, the segment is congruent to the altitude from $C$ of $\triangle ABC$.
As a result, we can calculate twice the area of the triangle in two ways to get
$$ c\cos A\cdot c \sin A = |AC||BC| = 2\,|\triangle ABC| = |AB||A'B| = c\cdot c\tan\phi \tag{$\star$}$$
which gives the result. $\square$

For a bit of a symbol-crunching solution, we can invoke the trigonometric from of Ceva's Theorem to write
$$\frac{\sin\phi}{\sin(A-\phi)}\cdot\frac{\sin\phi}{\sin(B-\phi)}\cdot\frac{\sin\phi}{\sin(90^\circ-\phi)}=1 \tag1$$
Thus,
$$\begin{align}
\sin^2\phi\tan\phi &= \sin(A-\phi)\sin(B-\phi) \tag2\\
&= \tfrac12\left(\cos((A-\phi)-(B-\phi)) - \cos((A-\phi)+(B-\phi))\right) \tag3\\
&= \tfrac12\left(\cos(2A-90^\circ) - \cos(90^\circ-2\phi)\right) \qquad (B=90^\circ-A) \tag4\\
&= \tfrac12\left(\sin2A-\sin2\phi\right) \tag5\\
&= \cos A\sin A - \cos\phi\sin\phi \tag6\\
&= \cos A\sin A - \cos^2\phi\tan\phi \tag7\\
(\sin^2\phi+\cos^2\phi)\tan\phi &= \cos A\sin A \tag8\\
\tan\phi &= \cos A\sin A \tag9
\end{align}$$
