Let $ABC$ be a triangle with incenter $I,$ and let $D,E,F$ be the midpoints of sides $BC, CA, AB$, respectively. Lines $BI$ and $DE$ meet at $P $ and lines $CI$ and $DF$ meet at $Q$. Line $PQ$ meets sides $AB$ and $AC$ at $T$ and $S$, respectively. Prove that $AS = AT$
Problem
Source: 2013 Saudi Arabia BMO TST I p6
Tags: geometry, incenter, midpoints, equal segments
rafaello
25.07.2020 11:54
This is kind of trivial.
Try to show that $B$, $Q$, $P$ and $C$ are concyclic with centre $D$. You can show that using fact that $DE$, $DF$ are midlines and $BI$ and $CI$ are the angle bisectors.
We will know that then $BC$ is the diameter, so $\angle BQC=\angle BPC = 90^\circ$.
We can also show that $B$, $I$, $T$ and $Q$ are concyclic and $C$, $I$, $P$ and $S$ are concyclic (that can be shown very easily), which implies that $A$, $T$, $S$ and $I$ are concyclic. Hence $\angle ASI=\angle ATI = 90^\circ$, thus $\triangle ASI \cong\triangle ATI$ and we are done.
WolfusA
25.07.2020 12:54
Theorem Given triangle $ABC$ with midpoints $D,F$ of sides $BC,AB$ respectively and it’s incenter $ I$. Let $M,N$ be points of tangency of incircle with sides $AB,AC$. Then lines $MN, DF, CI$ are concurrent.
By midline theorem $DF=\frac{AC}{2}$ and
$$DF\parallel AC\implies \measuredangle DQC= \measuredangle ACQ= \measuredangle QCD\implies DQ=CD.$$Observe
$$QF=|DQ-DF|=\frac{|BC-AC|}{2},\ MF=|BM-BF|=\left|\frac{AB+BC-AC}{2}-\frac{AB}{2}\right|=QF.$$Because $\measuredangle MFQ= \measuredangle BFD= \measuredangle MAN$ we have similarity in the same orientation of isosceles triangles
$$\triangle QFM\sim\triangle NAM\ (s,a,s)\implies \measuredangle FMQ= \measuredangle AMN.$$Since $M\in AF$ the equality proves $Q\in MN$. QED
By the theorem $P\in MN$ and analogously $Q\in MN$. Thus $(T,S)=(M,N)$ and we’re done by tangent segments theorem: $AM=AN$. QED
#1666, 261
dgreenb801
26.07.2020 04:19
We have $DF \parallel AC$, so $\angle DQC = \angle QCA = \angle QCD$. Thus, $QD=CD=BD$ (and by the same argument, these all also equal $PD$), so $\triangle DPQ$ is isoceles. Then $\angle ATS = \angle ATP = \angle TPD = \angle SQD = \angle ASQ = \angle AST$, so $AS=AT$.