Let $ABC$ be an obtuse triangle with $\angle A > \angle B > \angle C$, and let $M$ be a midpoint of the side $BC$. Let $D$ be a point on the arc $AB$ of the circumcircle of triangle $ABC$ not containing $C$. Suppose that the circle tangent to $BD$ at $D$ and passing through $A$ meets the circumcircle of triangle $ABM$ again at $E$ and $\overline{BD}=\overline{BE}$. $\omega$, the circumcircle of triangle $ADE$, meets $EM$ again at $F$. Prove that lines $BD$ and $AE$ meet on the line tangent to $\omega$ at $F$.
Problem
Source: KMO 2021 P6
Tags: geometry, circumcircle, tangent
mjk20016
13.11.2021 17:22
Let $Z$ be the intersection of $DE$ and $(ABC).$ From $\angle ADC=\angle ABM = \angle AEF = \angle ADF$ we have $(D, F, C)$ collinear. Since $DB$ is tangent to $\omega$ and $DB=EB$, $EB$ is also tangent to $\omega$. Now we have $\angle BED = \angle DAE$, so \[ \angle ZBE = \angle DEB - \angle BZD = \angle DAE - \angle DAB = \angle BAE \]gives us that $ZB$ is tangent to $(ABM).$ $\angle ZAC = \angle ZBC = \angle MAB$, so $AZ$ is symmedian with respect to $\triangle ABC$. \[ -1 = D(A, Z; B, C) = D(A, E;D, F) \]yields that $\square ADEF$ is a harmonic quadrilateral, then result follows.
mathlove_13520
14.11.2021 03:35
Other solution to the problem. Let $K, T$ each be the intersection of $AB$ and $(ABM)$, $BF$ and $(ADE)$. by angle chasing, we can easily see that $KF \parallel BC$ and $D, C, F$ collinear. From $F$, $FK, FB, FM, FC$ are harmonic pencils. On triangle $(ADE)$'s perspective, $\square DKTE$ is harmonic quadrilateral. Similarly, penciling from B, we get $\square ADEF$; a harmonic quadrilateral and now we are done.
KPBY0507
14.11.2021 20:33
A different solution with above I think.. Its trivial that $BE$ is tangent to $\omega$. Let $X$ be the intersection of $DE$ and the circumcircle of $\triangle{ABM}$. claim) $\triangle{AXB}$ is similar with $\triangle{AMC}$. <pf>
$\angle{XBA}=\angle{AEX}=\angle{AED}=180-\angle{ADB}=\angle{ACB}$.
Also, $\angle{AXB}=\angle{AMC}$.
Because of the claim, $\frac{XB}{AX}=\frac{MC}{AM}=\frac{MB}{AM}$, so $AXBM$ is homothetic. $-1=(B,A;X,M)=E(E,A;D,F) \rightarrow ADEF$ is homothetic!!