12. in the triangle $A B C$ let $J$ be the center of the semicircle that touches the side $B C$ in $A_{1}$ and the extensions of the sides $A C$ and $A B$ in $B_{1}$ and $C_{1}$ respectively. The line $A_{1} B_{1}$ intersects the line $A B$ at right angles in $D$. Let $E$ be the projection of $C_{1}$ onto the line $D J$. Determine the size of the angles $\angle B E A_{1}$ and $\angle A E B_{1}$.
## 1st solution

We first show that the points $C_{1}, E, C$ lie on a straight line by showing $C_{1} C \perp D J$. Let $K$ be the intersection of $J C$ and $A_{1} B_{1}$. Because $C B_{1}$ and $C A_{1}$ are tangent to the circle, $J C \perp A_{1} B_{1}$ and because $\angle K D C_{1}=\angle D C_{1} J=90^{\circ}$ $D C_{1} J K$ is a rectangle. Because $C J B_{1}$ is a right-angled triangle, we get

$$
J C_{1}^{2}=J B_{1}^{2}=J C \cdot J K=J C \cdot C_{1} D \quad \Rightarrow \quad \frac{J C_{1}}{J C}=\frac{C_{1} D}{C_{1} J}
$$

So the two right-angled triangles $C C_{1} J$ and $J D C_{1}$ are similar and we get $\angle C_{1} D J=\angle J C_{1} C$, from which $C_{1} C \perp D J$ follows. Thus the points $C_{1}, E, C$ lie on a straight line.

Because $\angle C A_{1} J=\angle C B_{1} J=\angle C E J=90^{\circ}$ the points $A_{1}, B_{1}$ and $E$ lie on the Thales circle above $C J$. We now have

$$
\angle D B A_{1}=90^{\circ}-\angle K A_{1} C=\angle A_{1} C J=180^{\circ}-\angle A_{1} E J=\angle D E A_{1}
$$

It follows that $B E A_{1} D$ is a chordal quadrilateral and we find $\angle B E A_{1}=90^{\circ}$. The quadrilateral $A D E B_{1}$ is also a chordal quadrilateral because $\angle E B_{1} A=\angle E J C=$ $\angle E D C_{1}$. This gives us $\angle A E B_{1}=\angle A D B_{1}=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_28_be699ce33ca02b2354f6g-13.jpg?height=1085&width=854&top_left_y=800&top_left_x=635)