Subfield: Geometry


Question: Let $\triangle A B C$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\angle A B C$ intersects $C D$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle A D E$ again at $F$. If $\angle A D F=45^{\circ}$, show that $C F$ is tangent to $\omega$.

Solution:
<img_3521>
Since $\angle C D F=90^{\circ}-45^{\circ}=45^{\circ}$, the line $D F$ bisects $\angle C D A$, and so $F$ lies on the perpendicular bisector of segment $A E$, which meets $A B$ at $G$. Let $\angle A B C=2 \beta$. Since $A D E F$ is cyclic, $\angle A F E=90^{\circ}$, and hence $\angle F A E=45^{\circ}$. Further, as $B F$ bisects $\angle A B C$, we have $\angle F A B=90^{\circ}-\beta$, and thus

$$
\angle E A B=\angle A E G=45^{\circ}-\beta, \quad \text { and } \quad \angle A E D=45^{\circ}+\beta \text {, }
$$

so $\angle G E D=2 \beta$. This implies that right-angled triangles $\triangle E D G$ and $\triangle B D C$ are similar, and so we have $|G D| /|C D|=|D E| /|D B|$. Thus the right-angled triangles $\triangle D E B$ and $\triangle D G C$ are similar, whence $\angle G C D=\angle D B E=\beta$. But $\angle D F E=\angle D A E=45^{\circ}-$ $\beta$, then $\angle G F D=45^{\circ}-\angle D F E=\beta$. Hence $G D C F$ is cyclic, so $\angle G F C=90^{\circ}$, whence $C F$ is perpendicular to the radius $F G$ of $\omega$.

It follows that $C F$ is a tangent to $\omega$, as required.
As $\angle A D F=45^{\circ}$ line $D F$ is an exterior bisector of $\angle C D B$. Since $B F$ bisects $\angle D B C$ line $C F$ is an exterior bisector of $\angle B C D$. Let $\angle A B C=2 \beta$, so $\angle E C F=(\angle D B C+\angle C D B) / 2=45^{\circ}+\beta$. Hence $\angle C F E=180^{\circ}-\angle E C F-\angle B C E-\angle E B C=180^{\circ}-\left(45^{\circ}+\beta+90^{\circ}-2 \beta+\beta\right)=45^{\circ}$. It follows that $\angle F D C=\angle C F E$, then $C F$ is tangent to $\omega$.
Note that $A E$ is diameter of circumcircle of $\triangle A B C$ since $\angle C D F=90^{\circ}$. From $\angle A E F=\angle A D F=45^{\circ}$ it follows that triangle $\triangle A F E$ is right-angled and isosceles. Without loss of generality, let points $A, E$ and $F$ have coordinates $(-1,0),(1,0)$ and $(0,1)$ respectively. Points $F, E$, $B$ are collinear, hence $B$ have coordinates $(b, 1-b)$ for some $b \neq-1$. Let point $C^{\prime}$ be intersection of line tangent to circumcircle of $\triangle A F E$ at $F$ with line $E D$. Thus $C^{\prime}$ have coordinates $(c, 1)$ and from $\overline{C^{\prime} E} \perp \overline{A B}$ we get $c=2 b /(b+1)$. Now vector $\overline{B C^{\prime}}=(2 b /(b+1)-b, b)=b /(b+1) \cdot(1-b, b+1)$, vector $\overline{B F}=(-b, b)=(-1,1) \cdot b$ and vector $\overline{B A}=(-(b+1),-(1-b))$. Its clear that $(1-b, b+1)$ and $(-(b+1),-(1-b))$ are symmetric with respect to $\overline{F E}=(-1,1)$, hence $B F$ bisects $\angle C^{\prime} B A$ and $C^{\prime}=C$ which completes the proof.
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Again $F$ lies on the perpendicular bisector of segment $A E$, so $\triangle A F E$ is right-angled and isosceles. Let $M$ be an intersection of $B C$ and $A F$. Note that $\triangle A M B$ is isosceles since $B F$ is a bisector and altitude in this triangle. Thus $B F$ is a symmetry line of $\triangle A M B$. Then $\angle F D A=\angle F E A=\angle M E F=45^{\circ}, A F=F E=F M$ and $\angle D A E=\angle E M C$. Let us show that $E C=C M$. Indeed,

$$
\begin{aligned}
\angle C E M & =180^{\circ}-(\angle A E D+\angle F E A+\angle M E F)=90^{\circ}-\angle A E D= \\
& =\angle D A E=\angle E M C .
\end{aligned}
$$

It follows that $F M C E$ is a kite, since $E F=F M$ and $M C=C E$. Hence $\angle E F C=\angle C F M=\angle E D F=45^{\circ}$, so $F C$ is tangent to $\omega$.
Let the tangent to $\omega$ at $F$ intersect $C D$ at $C^{\prime}$. Let $\angle A B F=$ $\angle F B C=\beta$. It follows that $\angle C^{\prime} F E=45^{\circ}$ since $C^{\prime} F$ is tangent. We have

$$
\frac{\sin \angle B D C}{\sin \angle C D F} \cdot \frac{\sin \angle D F C^{\prime}}{\sin \angle C^{\prime} F B} \cdot \frac{\sin \angle F B C}{\sin \angle C B D}=\frac{\sin 90^{\circ}}{\sin 45^{\circ}} \cdot \frac{\sin \left(90^{\circ}-\beta\right)}{\sin 45^{\circ}} \cdot \frac{\sin \beta}{\sin 2 \beta}=\frac{2 \sin \beta \cos \beta}{\sin 2 \beta}=1 .
$$

So by trig Ceva on triangle $\triangle B D F$, lines $F C^{\prime}, D C$ and $B C$ are concurrent (at $C$ ), so $C=C^{\prime}$. Hence $C F$ is tangent to $\omega$.