Let $D$ be a point inside an acute triangle $ABC$, such that $\angle BAD = \angle DBC$ and $\angle DAC = \angle BCD$. Let $P$ be a point on the circumcircle of the triangle $ADB$. Suppose $P$ are itself outside the triangle $ABC$. A line through $P$ intersects the ray $BA$ in $X$ and ray $CA$ in $Y$, so that $\angle XPB = \angle PDB$. Show that $BY$ and $CX$ intersect on $AD$.
Problem
Source: Switzerland - Swiss MO 2018 p4
Tags: geometry, concurrent, equal angles
Jalil_Huseynov
05.11.2021 18:49
First of all note that $(ADC)$ and $(ADB)$ are tangent to $BC$ at $C$ and $B$,respectively $\implies AD$ is $\text{A-median}$ in $\triangle ABC$ $(\star)$ Take any point $T$ on the ray $CB$ such that $B$ lies between $T$ and $C$.Then $\angle XPB = \angle PDB=\angle PBT$ $\implies$ $XY\parallel BC$ $(\star \star)$ From $(\star)$ and $(\star \star)$ we get $AD$,$BY$,$CX$ are concurrent $(\text{because of Ceva})$
Albert123
06.11.2021 05:43
just to explain some things of @above $ 1) AD $ is median because: Let $ AD \cap BC = Z $ then $ BZ ^ 2 = ZD.ZA = ZC ^ 2 \implies BZ = ZC $. $ 2) $ As $ XY // BC \implies \frac{AX}{XB} = \frac{AY}{YC} $ Then: $\frac{AX}{XB}.\frac{BZ}{ZC}.\frac{CY}{YA}=1$ Then by Ceva Normal: $ AD, BY, CZ $ are concurrent
HamstPan38825
29.06.2022 05:45
Because $\angle BAD = \angle DBC$, $\overline{BC}$ is tangent to $(BAD)$. Then, we may angle chase to obtain $$\angle XPB = \angle BDP = \angle PBC,$$which implies $\overline{XY} \parallel \overline{BC}$. Thus, by Ceva, it suffices to show $BE=EC$, where $E = \overline{AD} \cap \overline{BC}$. But this is trivial by radical axis, because $\overline{BC}$ is also tangent to $(ACD)$ by the second angle condition.