Let $O$ be the circumcenter of $\triangle ABC$. A circle with center $P$ pass through $A$ and $O$ and $OP$//$BC$. $D$ is a point such that $\angle DBA = \angle DCA = \angle BAC$. Prove that: Circle $(P)$, circle $(BCD)$ and the circle with diameter $(AD)$ share a common point.
Problem
Source: 2022 China Southeast Grade 11 P6
Tags: geometry
Behanorm_rifghcky
03.08.2022 10:30
let $BO\cap AC=E$,$CO\cap AB=F$,$K$ is the Miquel point of $AFBOCE$.Then just to verify that $K$ lies on the three circles.
amitsom
03.08.2022 10:53
$S=\odot R \cap \odot Q,T=SA\cap \odot R,DT$ passes $R$
$\angle TBA =\angle TCA=90^\circ -\angle A$
$\frac{Sin\angle BTA}{AB}=\frac{Sin\angle TBA}{AT}=\frac{Sin\angle TCA}{AT}=\frac{Sin\angle CTA}{AC}$
$\frac{BS}{SC}=\frac{Sin\angle BTA}{Sin\angle CTA}=\frac{AB}{AC}$
With $\angle BSC =\angle BDC=360^\circ-3\angle A$ we can now determine position of $S$
$\angle APO=180^\circ-2\angle AOP=2\angle C-2\angle B$(WLOG $AB>AC$)
need to proof $\angle ASO=\angle C-\angle B$
$E$ is on ray $OS$ makes $OA^2=OS\cdot OE$
$OB^2=OC^2=OS\cdot OE\Rightarrow \triangle OBS \sim \triangle OEB,\triangle OCS \sim \triangle OEC$
$\Rightarrow \frac{EB}{EC}=\frac{BS}{CS}=\frac{AB}{AC}$ and $\angle BEC=\angle A$
$\Rightarrow AE\perp BC,\angle ASO=\angle OAE=\angle C-\angle B$
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LoloChen
03.08.2022 11:24
amitsom wrote: 12345678 Thanks.
trinhquockhanh
20.04.2023 08:12
Firstly, we have a lemma: Go back to the main problem: