The incenter of \(\triangle ABC\) touches its sides at \(D,E,F\) respectively, and \(BE, CF\) intersect the incircle second time at \(M, N\) respectively. Extend \(MD\) to \(P\) so that \(DP=2MD\). Prove that \(DE\) is tangent to the circle \(\odot (DPN)\).

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Property1: \(\frac {DN}{EN}=\frac {DF}{EF}\).

Proof: Since \(CD,CE\) are tangent to the incircle, \(\triangle CDN\sim \triangle CFD\implies \frac {DN^2}{DF^2}=\frac {CN}{CF}\). Symmetrically we have \(\frac {EN^2}{EF^2}=\frac {CN}{CF}\), thus \(\frac {DN}{DF}=\frac {EN}{EF}\) and switching \(DF,EN\) gives us our desired equality.

Property2: Let \(L\) denote the midpoint of \(EM\), then \(\triangle DML\sim \triangle DEF\)

Proof: Note that \(\angle ILB=\angle IFB=\angle IDB=90^{\circ}\), therefore \(D,I,L,F,B\) are concyclic where \(I\) is the incenter of \(ABC\). Hence \(\angle DLM=\angle DFB=\angle DEF\). Since \(\angle DML=\angle DFE\) by cyclic, we have our desired similarity by AA

Correlary: \(\frac {DN}{EN}=\frac {DF}{EF}=\frac {DM}{ML}=\frac {2DM}{EM}\)

Main proof: It suffices to show \(\angle DPN=\angle NDE=\angle NME\), since \(\angle PDN=\angle NEM\) by cyclic, we will prove that \(\triangle NDP\sim \triangle NEM\). This is equivalent to \(\frac {DN}{EN}=\frac {DP}{EM}=\frac {2DM}{EM}\) by SAS similarity, which follows from the correlary above.

Note: A cyclic quadrilateral that satisfies property 1 is called harmonic quadrilateral, and it is intimately related to concepts such as harmonic division and poles and polar. – Xuming Liang · 1 year, 1 month ago

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– Alan Yan · 1 year, 1 month ago

Btw do you have a document that goes over some incircle properties? I am not that familiar with them and they will be very useful.Log in to reply

– Xuming Liang · 1 year, 1 month ago

BTW did u get this from AoPS? I just went on and saw the exact same problemLog in to reply

– Alan Yan · 1 year, 1 month ago

Yes I did :)Log in to reply