The tangents at $A$ and $B$ to the circumcircle of the acute triangle $ABC$ intersect the tangent at $C$ at the points $D$ and $E$, respectively. The line $AE$ intersects $BC$ at $P$ and the line $BD$ intersects $AC$ at $R$. Let $Q$ and $S$ be the midpoints of the segments $AP$ and $BR$ respectively. Prove that $\angle ABQ=\angle BAS$.
Problem
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Tags: geometry, circumcircle, geometry proposed
dgreenb801
17.01.2011 01:03
Where is $Q$? And what is the point of having point $S$ if you could just say $\angle BAP$ instead of $\angle BAS$?
WakeUp
17.01.2011 02:14
dgreenb801 wrote: Where is $Q$? And what is the point of having point $S$ if you could just say $\angle BAP$ instead of $\angle BAS$? Sorry about this, the mistake is not my own as this is the exact wording of the problem. However http://www.imomath.com/ believes that Q is the midpoint of AP and S is the midpoint of BR. I've just done a quick sketch and this seems right, so i've edited the problem.
Luis González
17.01.2011 02:31
Oh my Lord, it's this problem again !. http://www.artofproblemsolving.com/viewtopic.php?t=19806 http://www.artofproblemsolving.com/viewtopic.php?t=207440 http://www.artofproblemsolving.com/viewtopic.php?f=46&t=275380
cwein3
29.08.2011 11:31
Very nice problem It is well known that $AP$ and $BR$ are symmedians of the triangle $ABC$. See here: http://web.mit.edu/yufeiz/www/olympiad/geolemmas.pdf Let the midpoint of $BC$ be $X$ and the midpoint of $AC$ be $Y$. Let $AX$ intersect the circumcircle again at $Z$ and $BY$ intersect the circumcircle at $W$. Note that $\triangle BAP \sim \triangle ZAC$, and $\triangle ARB \sim \triangle WCB$, from trivial angle chasing using the fact that $AP$ and $BR$ are symmedians. Then we must have $\angle AZR = \angle ABQ$ and $\angle XWB = \angle SAB$ by similar triangles. But since $\angle XRB = \angle WBA = \angle WZA$, $WRXZ$ is cyclic, so $\angle SAB = \angle XWB = \angle AZR = \angle ABQ$, as desired.