
1. Let $A B C$ be a triangle with $\angle B A C>90^{\circ}$. Let $D$ be a point on the segment $B C$ and $E$ be a point on the line $A D$ such that $A B$ is tangent to the circumcircle of triangle $A C D$ at $A$ and $B E$ is perpendicular to $A D$. Given that $C A=C D$ and $A E=C E$, determine $\angle B C A$ in degrees.


Solution: Let $\angle C=2 \alpha$. Then $\angle C A D=\angle C D A=90^{\circ}-\alpha$. Moreover $\angle B A D=2 \alpha$ as $B A$ is tangent to the circumcircle of $\triangle C A D$. Since $A E=A D$, it gives $\angle A E C=2 \alpha$. Thus $\triangle A E C$ is similar to $\triangle A C D$. Hence

$$
\frac{A E}{A C}=\frac{A C}{A D}
$$

But the condition that $B E \perp A D$ gives $A E=A B \cos 2 \alpha=c \cos 2 \alpha$. It is easy to see that $\angle B=90^{\circ}-3 \alpha$. Using sine rule in triangle $A D C$, we get

$$
\frac{A D}{\sin 2 \alpha}=\frac{A C}{\sin (90-\alpha)}
$$

This gives $A D=2 b \sin \alpha$. Thus we get

$$
b^{2}=A C^{2}=A E \cdot A D=(c \cos 2 \alpha) \cdot 2 b \sin \alpha
$$

Using $b=2 R \sin B$ and $c=2 R \sin C$, this leads to

$$
\cos 3 \alpha=2 \sin 2 \alpha \cos 2 \alpha \sin \alpha=\sin 4 \alpha \sin \alpha
$$

Writing $\cos 3 \alpha=\cos (4 \alpha-\alpha)$ and expanding, we get $\cos 4 \alpha \cos \alpha=0$. Therefore $\alpha=90^{\circ}$ or $4 \alpha=90^{\circ}$. But $\alpha=90^{\circ}$ is not possible as $\angle C=2 \alpha$. Therefore $4 \alpha=90^{\circ}$ which gives $\angle C=2 \alpha=45^{\circ}$.
