# 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!

• It may be helpful that $\triangle OBC$ is a right-angled triangle. Jul 9, 2021 at 21:51

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*}

• Thank you for the response! Why can we work with $AB=1$? Jul 9, 2021 at 22:26
• @Medi the angles are still equal by scaling the whole diagram until $AB=1$. Otherwise you can also write $OB = \frac{\sin\varphi}{\cos\alpha} AB$ and $BC = AB\sin \alpha$, and after cancelling the scaling factor $AB$ the proof would still hold. Jul 9, 2021 at 22:29

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}

• Thank you for the response! I am not sure I see how $A'B=c\tan\varphi$. Jul 10, 2021 at 21:42
• @Medi: $$\text{tan} = \frac{\text{opposite}}{\text{adjacent}} \quad\to\quad \text{opposite} = \text{adjacent}\cdot\text{tan}$$
– Blue
Jul 10, 2021 at 21:51
• But I don't see angle $\varphi$ in this triangle. Jul 10, 2021 at 21:52
• @Medi: Angle $\phi$ is at vertex $A$ in $\triangle ABA'$.
– Blue
Jul 10, 2021 at 21:54
• Silly me, thank you! A really nice approach and idea. Jul 10, 2021 at 22:03