Let $\triangle ABC$ be an acute-angled triangle. Let $P$ be the midpoint of $BC$, and $K$ the foot of the altitude from $A$ to side $BC$. Let $D$ be a point on segment $AP$ such that $\angle BDC = 90^\circ$. Let $E$ be the second point of intersection of line $BC$ with the circumcircle of $\triangle ADK$. Let $F$ be the second point of intersection of line $AE$ with the circumcircle of $\triangle ABC$. Prove that $\angle AFD = 90^\circ$.
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Tags: TST, Chile, geometry
TestX01
23.10.2024 01:50
Note we just want $(AFD)$ tangent to $ED$, as $\angle EDA=\angle EKA=90^\circ$ by Bowtie. But this is true as $EA\cdot EF=EB\cdot EC=ED^2$ as $ED$ tangent to $(BDC)$ by definition of tangency $P$ is centre.
anirbanbz
29.10.2024 01:41
$\textbf{Claim:}$ $ED$ $\text{is tangent to}$ $(BDC)$. $\newline$ $\textbf{Proof:}$ $(BDK)$ $\text{is clearly the circle with center}$ $P$ $\text{and radius}$ $r=PB=PC$. $\text{Moreover,}$ $\angle ADE = \angle AKE = 90^\circ$ $\text{, and}$ $\angle ADB = 180^\circ -BDP = 180^\circ - (\frac{180^\circ -\angle DPB}{2})= 180^\circ - (90^\circ - \frac{\angle DPB}{2})=90^\circ +\angle DCB$. $\newline$ $\text{Combining we get:}$ $\angle EDB=\angle ADB - \angle ADE= 90^\circ + \angle DCB - 90^\circ=\angle DCB$. $\newline$ $\Longrightarrow$ $\text{Thus, the claim is proven.}$ $\newline$ $\textbf{Finishing:}$ $\text{By Radical axis theorem,}$ $ED$ $\text{must also be tangent to}$ $(AFD)$ $\text{at}$ $D$$\text{, and thus,}$ $\angle AFD = 180^\circ - \angle ADE = 90^\circ$$\text{, which is what we wanted.}$ $\blacksquare$