Let $ABC$ be a triangle such that $AB=AC$ and $\angle BAC$ is obtuse. Point $O$ is the circumcenter of triangle $ABC$, and $M$ is the reflection of $A$ in $BC$. Let $D$ be an arbitrary point on line $BC$, such that $B$ is in between $D$ and $C$. Line $DM$ cuts the circumcircle of $ABC$ in $E,F$. Circumcircles of triangles $ADE$ and $ADF$ cut $BC$ in $P,Q$ respectively. Prove that $DA$ is tangent to the circumcircle of triangle $OPQ$.
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Tags: geometry, circumcircle, geometric transformation, reflection
P2nisic
22.05.2022 16:43
Let $D'$ be the reflection of $D$ over $OA$ and $X=(ABC)\cap AD'$ Perfor a $\sqrt{bc}$ inversion then: $D\leftrightarrow X$ and $M\leftrightarrow O$ Let $E,F$ go to $E',F'$ which are obvioysly on $BC$ $A,E',O,F',X$ are all on the same circle because $D,E,M,F$ are colinear. Now $<EAF=<EAM+<MAF=<MAE'+<MAF'=F'AE'$ and $AF.AF'=AE.AE'\Rightarrow \frac{AE}{AF'}=\frac{AF}{AE'}$ so $AEF\approx AF'E'$ For spiral similaraty we have $E,F',X$ colinear and $F,E',X$ collinear. Now from inversion we have: $XF'E\leftrightarrow (DAE'F)$ and $XE'F\leftrightarrow (DAF'E)$ Which mean that $P\equiv F',Q\equiv E'$ $<XF'O=<XAO=<XAM=<AME\Rightarrow P,O,M,E$are on the same circly similar for $Q,M,O,F$. $<DAP=<PEM=<POA$ which means that $DA$ is tangent to the circumcircle of triangle $OPQ$
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ismayilzadei1387
28.03.2024 19:48
it is quite boring angle chasing after some angle chasing $O$ lies on polar of $D$ with phantom point it is obvious that $(APQ) \cap BC$ at $O$