Outside the square $ABCD$, a semi-circle of diameter $BC$ is constructed. On the arc $BC$, point $L$ is marked. If $AL=2(BL)$, calculate $m \sphericalangle ALC$. (Answer: $60^\circ$)
I made the drawing and got a solution by trigonometry. Does anyone have a solution using geometry?
$x= BL$, $AL = 2x$
$α,β$ : $\sphericalangle LBC, \sphericalangle ALC$
$\triangle CLB_{(ret.)}\\
\angle BCL = 90-\alpha \implies \angle ACL =135^o +\alpha\\
\angle ACB = 90-\beta \implies \angle BAL = \beta – \alpha$
$\begin{array} :\triangle ACL:\dfrac{l\sqrt{2}}{\sin\beta}=\dfrac{2x}{\sin(135^{\circ}-\alpha)}(I)\\\triangle LAB:\dfrac{2x}{\sin(90^{\circ}+\alpha)}=\dfrac{x}{\sin(\beta-\alpha)}=\dfrac{l}{\sin(90^{\circ}-\beta)}\end{array}$
$\dfrac{2x}{\sin(90^o+α)}=\dfrac{l}{\sin(90^o−β)}
⟺
2x=\dfrac{l\cos α}{\cos β}\\
sen(135^o−α)=\frac{\sqrt2(\cos α+\sin α)}{2}\\
\text{From (I) : } \dfrac{l\sqrt2}{\sin β}=\dfrac{l\cos α}{\dfrac{\sqrt2}{2}(\cos α+\sin α)\cos β}\\
\therefore\ 1+\tan α=\tan β\ (∗)\ . \\
\dfrac{2x}{\sin(90^o+α)}=\dfrac{x}{\sin(β−α)}
\\\qquad
\implies 2(\sin β \cos α−\cos β \sin α)= \cos α\qquad(\div \cos\alpha \cos\beta)
\\\qquad
\implies\tan β−\tan α=\dfrac{1}{2\cos β}\\
\text{From $(∗)$ : }1=\dfrac{1}{2\cos β}\ ⟹\ \cos β=\dfrac{1}{2} \therefore \boxed{β=60^\circ }
$
(Solution by ani_pascual)
Best Answer
Build an equilateral triangle $ABE$ as in the picture. Let $F$ be the midpoint of $BC$. Let $G$ be the reflection of $B$ in $EF$. Note that $FG = FB = FC$, hence $G$ lies on the circle with diameter $BC$.
Note that $A,B,G$ lie on a circle with center $E$ and radius $AE=BE=GE$. Therefore $\angle AGB = \frac 12 \angle AEB = 30^\circ$. Furthermore, by symmetry, $\angle EGF = \angle FBE = 90^\circ - 60^\circ = 30^\circ$.
Moreover, $\angle FEG = \frac 12 \angle BEG = \angle BAG$. This shows that $\triangle EFG \sim \triangle ABG$ by AAA. Since $EG = EB = AB = BC = 2BF = 2FG$, it follows that $AG=2BG$.
Since $G$ lies on the circle with diameter $BC$ and satisfies $AG=2BG$, it follows that $G$ coincides with $L$.
To finish, simply note that $\angle CLA = \angle CGA = 90^\circ - 30^\circ = 60^\circ$.