Let $\Omega $ be the circumcincle of the quadrilateral $ABCD $ , and $E $ the intersection point of the diagonals $AC $ and $BD $ . A line passing through $E $ intersects $AB $ and $BC$ in points $P $ and $Q $ . A circle ,that is passing through point $D $ , is tangent to the line $PQ $ in point $E $ and intersects $\Omega$ in point $R $ , different from $D $ . Prove that the points $B,P,Q,$ and $R $ are concyclic .
Problem
Source: Moldova TST 2018
Tags: geometry
06.04.2018 15:52
06.04.2018 17:59
FedeX333X wrote: Nice problem!
It's not a hint, actually it's what we're supposed to Prove in order to use the Miquel point properties.
09.09.2020 14:22
$\angle PER=\angle BDR=\angle BAR=\angle PAR$. So $ARPE$ is cyclic. $\angle QER=\angle PER=\angle BAR=\angle BCR=\angle QCR$. So $QREC$ is cyclic too. $\angle RQB=\angle RQC=180-\angle REC=\angle AER=\angle APR$.
09.09.2020 15:44
MarkBcc168 wrote:
How does the conclusion follow that $E$ is the miquel point ?can anyone please help me
21.11.2021 07:45
yayitsme wrote: MarkBcc168 wrote:
How does the conclusion follow that $E$ is the miquel point ?can anyone please help me In the $ABCEQP$, $R \in \odot (ECR), \odot (AEP) \implies R=\text{Miquel Point}$ this is since chasing angles.
12.03.2023 10:55
$ERQC$ & $EARP$ are cyclic by angle chasing