In the acute-angled triangle $ABC$, $P$ is the midpoint of segment $BC$ and the point $K$ is the foot of the altitude from $A$. $D$ is a point on segment $AP$ such that $\angle BDC=90$. Let $(ADK) \cap BC=E$ and $(ABC) \cap AE=F$. Prove that $\angle AFD=90$.
Problem
Source: Turkey JBMO TST 2024 P1
Tags: geometry
13.05.2024 10:23
Let the circle with diameter $AD$ (which we will denote by $(AD)$ ) intersect $(ABC)$ at $F'$. Its easy to see that the $(BDC)$ and $(AD)$ are tangent at $D$, because their centers are collinear on the line $AP$. By radical axis on $(ABC), (AD), (BDC)$, we see that $BC, AF'$ and the line through $D$ perpendicular to $AP$ intersect at one point, say $E'$. Then, we have $\angle ADE' = \angle AKE' = 90$ so $ADKE'$ is concyclic. This solves the problem.
13.05.2024 20:23
Nice and straightforward. Since $\angle ADE = \angle AKE = 90^{\circ}$, it suffices to show $ED^2 = EF \cdot EA$. But $EF \cdot EA = EB \cdot EC$ from the circumcircle of $ABC$, so it suffices to justify $ED^2 = EB \cdot EC$, equivalent to $\angle EDB = \angle BCD$. But if $\angle BCD = x$, then $\angle BPD = 2x$ and $\angle BDP = 90^{\circ} - x$ as $DP = BP = PC$, thus $\angle EDB = 180^{\circ} - \angle ADE - \angle BDP = x$, as desired.
21.05.2024 10:43
nice problem
25.09.2024 17:39
$BP=PC=DP$. Then Pythagoras in $\triangle EDP$ gives $$ED^2+BP^2=EP^2=(EB+BP)^2$$Thus $$ED^2=EB(EB+2BP)=EB\cdot EC=EF\cdot EA$$as desired.