Determine an angle in two overlapping triangles

Question

A triangle $ABC$ is given as shown below. We know that $i = i_1$ and $k = k_1 = k_2$. Determine angle $\gamma$ $geometrically$. Note: Through variations in 'geogebra' I think that $\gamma = \frac{3}{4} \pi $ (or 135°). But how prove it? Does knowing $tan(\alpha)= \frac{k}{i+k} ,$ and $ tan(\beta)=\frac{i}{i+2k} $ help?

Answer 1

Denote the two unnamed bottom points by $D$ and $E$, as in the picture. Let $\omega$ be the circle with diameter $DR$ and $\Omega$ be the circle with diameter $DB$. Let $C'\neq D$ be the second intersection point of $\omega$ and $\Omega$.

Note that $\angle RC'D = 90^\circ = \angle DC'B$, hence $C'$ lies on $RB$.

Also note that $\angle AC'D = \angle ARD = 45^\circ$ and $\angle DC'S = \angle DC'B + \angle BC'S = 90^\circ + \frac 12 \angle BES = 90^\circ + 45^\circ = 135^\circ$. Therefore $\angle AC'D + \angle DC'S = 180^\circ$, hence $C'$ lies on $AS$.

This shows that $C'$ is the intersection of $BR$ and $AS$, hence $C'=C$. To finish, note that $\angle ACB = \angle ACD + \angle DCB = 45^\circ+90^\circ = 135^\circ$.

Answer 2

Yes, having remarked that $\tan(\alpha)= \frac{k}{i+k}$ and $\tan(\beta)= \frac{i}{i+2k}$ is very useful because :

$$\tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan\beta}=\frac{\frac{k}{i+k}+\frac{i}{i+2k}}{1-\frac{k}{i+k}\frac{i}{i+2k}}=\frac{i^2+2ik+2k^2}{i^2+2ik+2k^2}=1 \ \text{const.}$$

Therefore, using the fact that the sum of angles in triangle $ABC$ is equal to $\pi$, we have:

$$\tan \gamma = \underbrace{\tan(\pi-(\alpha+\beta))}_{-\tan(\alpha+\beta)} = -1.$$

... a constant... **which is indeed equal to $\tan(\underbrace{3 \pi/4)}_{135°}=-1$.

(As a consequence, $C$ belongs to an arc of circle passing through $A$ and $B$ subtanding angle $135°$).