Suppose that $ ABCD$ is a convex quadrilateral. Let $ F = AB\cap CD$, $ E = AD\cap BC$ and $ T = AC\cap BD$. Suppose that $ A,B,T,E$ lie on a circle which intersects with $ EF$ at $ P$. Prove that if $ M$ is midpoint of $ AB$, then $ \angle APM = \angle BPT$.
Problem
Source: Iran TST 2004
Tags: geometry, circumcircle, ratio, projective geometry, geometry proposed
Erken
09.01.2009 18:21
What is $ P$?
Omid Hatami
09.01.2009 19:23
$ P$ is intersection point of circle passing through $ A,B,T,E$ with $ EF$.
April
10.01.2009 02:09
We have $ E(DTCF)$ is a harmonic bundles, so by intersecting them with the circumcircle of quadrilateral $ (EATB)$ we conclude that $ ATBP$ is a harmonic quadrilateral. It implies $ PT$ is the symmedian line of triangle $ ABP$, i.e., $ \angle APM=\angle BPT$.
TheBernuli
18.02.2014 22:12
How do you define harmonic points if these points aren't on the same line?
fmasroor
20.02.2014 06:27
Harmonic quadrilaterals: Cyclic quadrilaterals with the length ratio properties of the normal harmonic division. They have a slew of cool properties, some of them look similar to the normal harmonic division ones.
sayantanchakraborty
09.09.2014 19:31
My solution: Let $ET \cap AB=Y,ET \cap DC=X$.As $EX,DB,AC$ are concurrent cevians in $\triangle{EDC}$ we have $(F,Y;A,B)=-1$.Thus $FA \cdot FB=FY \cdot FM=FP \cdot FE \implies PEMY$ cyclic.So $\angle{PEY}=\angle{PMY}=\angle{PMA}$.Also note that $\angle{PEA}=\angle{PBA}=\angle{PBY}$.So $\angle{BPM}=\angle{PMA}-\angle{PBY}=\angle{PEY}-\angle{PEA}=\angle{AET}=\angle{APT}$ or $\angle{BPT}=\angle{APM}$ as desired.Also as a note of interest note that $PATB$ is a harmonic quadrilateral.