Let $A$ and $B$ be two points on the circle with diameter $[CD]$ and on the different sides of the line $CD.$ A circle $\Gamma$ passing through $C$ and $D$ intersects $[AC]$ different from the endpoints at $E$ and intersects $BC$ at $F.$ The line tangent to $\Gamma$ at $E$ intersects $BC$ at $P$ and $Q$ is a point on the circumcircle of the triangle $CEP$ different from $E$ and satisfying $|QP|=|EP|. \: AB \cap EF =\{R\}$ and $S$ is the midpoint of $[EQ].$ Prove that $DR$ is parallel to $PS.$
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Tags: geometry, circumcircle, geometry proposed
lssl
15.12.2010 13:09
Hi , crazyfehmy , where is $E$ , I can't understand.
crazyfehmy
15.12.2010 13:22
lssl wrote: Hi , crazyfehmy , where is $E$ , I can't understand. Ok, let me explain where it is. $E$ is the point of intersection of $\Gamma$and $[AC]$ and $E$ is different from $A$ and $C.$
RaleD
26.12.2010 20:49
crazyfehmy wrote: $AP \cap EF =\{R\}$ $ It should be AB
RaleD
26.12.2010 21:44
It is harder to draw the diagram than to solve the problem. By angle chasing we get $\angle FPS=90-\angle CFE$. Let K be intersection of $DR$ and $FC$. We know that B, R and A are collinear, because of Simpson's line theorem. So $\angle FKD=90-\angle CFE$
FantasyLover
26.12.2010 22:23
First of all, we claim that $Q\in EF$. Let $Q'$ be the second intersection of $EF$ and $(CPE)$. We prove that $\angle PEQ'=\angle PQ'E$.
Indeed, if $K$ is an arbitrary point on line $PE$ such that $E$ is between $P$ and $K$, we have the following:
$\angle PEQ'=180^{\circ}-\angle FEK=180^{\circ}-\angle FCE=\angle PCE=\angle PQ'E$. Hence, $Q\equiv Q'$, and $Q$ lies on $EF$.
Returning to the problem, we prove that $DR\parallel PS$ by proving that $\angle PSR=\angle DRS$.
However, $\angle PSR=\angle PSQ=90^{\circ}$, and it suffices to prove that $\angle DRF=180^{\circ}-\angle DRS$ is equal to $90^{\circ}$.
Now, $\angle RFD=\angle EFD=\angle ECD=\angle ACD=\angle ABD=\angle RBD$, so quadrilateral $RFBD$ is cyclic.
Hence, $\angle DRF=180^{\circ}-\angle DBC=90^{\circ}$ since $CD$ is diameter of $(CBD)$, and we are done. $\blacksquare$
crazyfehmy
26.12.2010 23:16
Yes, you are right, it should be $AB \cap EF =\{R\}$ [moderator edit: fixed ]
sunken rock
23.04.2011 20:17
Let $Q'=EF\cap (CEP)$; from $\angle CFE=\angle CEP=\angle CQP$ it follows that $PQ'$ is tangent to $(QFC)$, i.e. $Q'P^2=PC\cdot PF=PE^2$, or $Q'\equiv Q$, hence $Q, E, F$ are collinear, so we need to show $DR\perp EF$, but $A, R, B$, are on the Simson line of $D$ w.r.t $\Delta FEC$, done. Best regards, sunken rock
ehuseyinyigit
23.10.2024 12:33
An easy angle chase even though couldn't realized the Simson configuration $----------------------------------------------------$ Our claim is that points $A$, $D$, $E$ and $R$ are concyclic. Since $$\angle PQC=\angle PEC=\angle CFD=\angle CDE \quad \text{and} \quad \angle CBA=CBD$$we have $\angle ARE=ADE$ which implies the points $A$, $D$, $E$ and $R$ being concyclic. Thus, $\angle DRE=\angle PSR\Longleftrightarrow DR||PS$ as desired.
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