The incenter of △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 ⊙(DPN).
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More incircle properties!
Property1: ENDN=EFDF.
Proof: Since CD,CE are tangent to the incircle, △CDN∼△CFD⟹DF2DN2=CFCN. Symmetrically we have EF2EN2=CFCN, thus DFDN=EFEN and switching DF,EN gives us our desired equality.
Property2: Let L denote the midpoint of EM, then △DML∼△DEF
Proof: Note that ∠ILB=∠IFB=∠IDB=90∘, therefore D,I,L,F,B are concyclic where I is the incenter of ABC. Hence ∠DLM=∠DFB=∠DEF. Since ∠DML=∠DFE by cyclic, we have our desired similarity by AA
Correlary: ENDN=EFDF=MLDM=EM2DM
Main proof: It suffices to show ∠DPN=∠NDE=∠NME, since ∠PDN=∠NEM by cyclic, we will prove that △NDP∼△NEM. This is equivalent to ENDN=EMDP=EM2DM 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.
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More incircle properties!
Property1: ENDN=EFDF.
Proof: Since CD,CE are tangent to the incircle, △CDN∼△CFD⟹DF2DN2=CFCN. Symmetrically we have EF2EN2=CFCN, thus DFDN=EFEN and switching DF,EN gives us our desired equality.
Property2: Let L denote the midpoint of EM, then △DML∼△DEF
Proof: Note that ∠ILB=∠IFB=∠IDB=90∘, therefore D,I,L,F,B are concyclic where I is the incenter of ABC. Hence ∠DLM=∠DFB=∠DEF. Since ∠DML=∠DFE by cyclic, we have our desired similarity by AA
Correlary: ENDN=EFDF=MLDM=EM2DM
Main proof: It suffices to show ∠DPN=∠NDE=∠NME, since ∠PDN=∠NEM by cyclic, we will prove that △NDP∼△NEM. This is equivalent to ENDN=EMDP=EM2DM 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.
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BTW did u get this from AoPS? I just went on and saw the exact same problem
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Yes I did :)
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.