Nice Geometry Problem

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 $$\odot (DPN)$$.

Note by Alan Yan
2 years, 7 months ago

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More incircle properties!

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.

- 2 years, 7 months 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.

- 2 years, 7 months ago

BTW did u get this from AoPS? I just went on and saw the exact same problem

- 2 years, 7 months ago

Yes I did :)

- 2 years, 7 months ago