Let $ABC$ be a triangle, $\Gamma$ its circumcircle and $l$ the tangent to $\Gamma$ through $A$. The altitudes from $B$ and $C$ are extended and meet $l$ at $D$ and $E$, respectively. The lines $DC$ and $EB$ meet $\Gamma$ again at $P$ and $Q$, respectively. Show that the triangle $APQ$ is isosceles.
Problem
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Tags: geometry, circumcircle
hectorraul
19.06.2019 17:16
Hint: $PQ\parallel ED$
a_simple_guy
19.06.2019 18:48
Let $J$ and $K$ be the foot of the perpendicular from $C$ and $B$ to $AB$ and $AC$ respectively. So we have
$\angle EAB=\angle ACB=\angle AQB=\angle AJK$. So if $AQ \cap JK=F$ then $JFBQ$ lies on a circle. Hence $AFK=\angle ABQ=\angle APQ$. Let $EQ \cap DP=G$. Note that $\angle JEA=\angle KBC=\angle DBC=90-\angle C \implies D,E,B,C$ lies on a circle. So $\angle GBA=\angle GDA=\angle GPQ \implies PQ \parallel DE$. Hence $\angle AFK =\angle AQP$. Hence $AQ=AP$.
GorgonMathDota
19.06.2019 20:01
Notice that proving $AP = AQ$ is equivalent to proving $\angle ABP = \angle ACQ$, which is equivalent to proving $DEBC$ is cyclic. It suffices to prove that $\angle EDB = \angle ECB$. But, \[ \angle EDB = 90^{\circ} - \angle DAC = 90^{\circ} - \angle B = \angle ECB \]We are hence finished.
Packito
01.08.2019 22:24
This problem was proposed by me ( Nicolas De la Hoz from Colombia)
Pluto1708
02.08.2019 01:09
Let $O$ be the circumcenter of $\odot{ABC}$.Then it suffices to show $OA\perp KL \Rightarrow KL \parallel DE$ since $DE$ is antiparallel to $BC$.But it's easy to see by angle chssing that $DEBC$ is cyclic. Hence by reim we are done. $\blacksquare$
bever209
11.08.2020 01:34
@3above, how is the angle condition equivalent to proving DEBC is cyclic?
AlexCenteno2007
29.11.2024 22:03
Sean M and N First we have the intersections of BD with AC and EC with AB. Then we consider the ∠ACB=β y ∠ABC=α. Because you are ∠EAB Semi-registered, we have to ∠EAB=∠ACB=β. Similarly, we have that the angle CAD=α So we have to ∠ANE=90°, We have to ∠AEN=90-β And similarly, ∠ADM=90-=α. Therefore, the EDCB quadrilateral is cyclical since ∠ECB=∠EDB. Later ∠EBD=∠ECD And as the BNMC quadrilateral is also cyclical(Since the angle BNC=to the angle BMC=90°) We have to ∠EBA=∠DCA, That is, ∠QBA=∠ACP. Hence, AQ=AP, from where ∠APQ=∠AQP. Therefore we have concluded that the triangle APQ is isosceles