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
25.07.2020 11:54
This is kind of trivial.
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 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
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 $AS=AT$.