Let $ABC$ be an obtuse-angled triangle with $ \angle ABC>90^{\circ}$, and let $(c)$ be its circumcircle. The internal angle bisector of $\angle BAC$ meets again the circle $(c)$ at the point $E$, and the line $BC$ at the point $D$. The circle of diameter $DE$ meets the circle $(c)$ at the point $H$. If the line $HE$ meets the line $BC$ at the point $K$, prove that: (a) the points $K, H, D$ and $A$ are concyclic (b) the line $AH$ passes through the point of intersection of the tangents to the circle $(c)$ at the points $B$ and $C$.
Problem
Source: Cyprus 2022 TST-2 Problem 3
Tags: geometry, symmedian
lgkarras
21.02.2022 21:26
https://mathematica.gr/forum/viewtopic.php?f=58&t=71185#p345811
aleksijt
22.02.2022 02:11
For (a), just $\angle KHA = \angle ACE = \angle ACB + \angle EAC = \angle KDA$. Let $O$ be the circumcenter of this circle. Then $O$ is the midpoint of $KD$ (and so is on $BC$), and is tangent to $(c)$ at $A$. Because $OA=OH$ and $H \in (c)$, $OH$ is also tangent to $(c)$ at $H$. In other words (w.r.t. $(c)$), the pole of $AH$ lies on $BC \Leftrightarrow$ the pole of $BC$ lies on $AH$. Hence, (b) follows.
Ibrahim_K
26.11.2022 18:35
Another solution a) Let $\angle DAC = a , \angle DCA= b$ then $\angle ADB = a+b = \angle ADK$ also $\angle DCE=a \implies \angle ACE=a+b \implies \angle AHK=a+b$ Hence $\angle AHK= \angle ADK \implies ADHK$ cyclic. b)Let $M$ be midpoint $BC$.Obviously $EB=EC \implies EM \bot BC \implies M$ lies on $(DHE)$ We have $\angle AHD=90-a-b \implies \angle AKD=90-a-b =\angle AKM(1)$ On the other hand $\angle EDM=a+b \implies \angle DEM=90-a-b=\angle AEM(2)$ By $1$ and $2$ $AKEM$ cyclic Let $\angle BAH =a-c \implies \angle HAD=c \implies \angle HKD=c \implies \angle EKM=c \implies \angle EAM=c \implies \angle MAC=a-c$ Thus $AH$ is $A-$symmedian $\implies$ $AH$ passes through the point of intersection of the tangents to the circle $(c)$ at the points $B$ and $C$. so we are done