Triangle ABC has an incircle $(I)$ which contacts $BC,CA,AB$ at $D,E,F$. Let $BP,CQ$ be bisectors of $\angle ABC,\angle ACB$ ($P \in AC,Q\in AB$). Line $AI$ intersects circle $(I)$ at $J$ (point $J$ is between two points $A$ and $I$). Two new lines through point $A$ (which is parallel to $JE,JF$) cut $DE,DF$ at $M,N$.$ID$ intersects $PQ$ at $R$. Show that if $AR$ intersects midperpendicular of $MN$ at $X$, then $X\in(I)$.

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I think a lot about this interesting problem. maybe the shortest way to solve this is use theorem about complete quadrilateral. But it needs some another factors to solve. Help me to show this please!


We prove a series of lemmas:

Lemma 1: AW,AD are isogonal in $\angle BAC$

Proof: Let $X\equiv KL \cap BC$ (notice $X$ is on the external angle bisector of $\angle BAC$) and let $W'$ be the isogonal conjugate of $W$ then: $$(X,D;B,C) \stackrel{I}{=} (X,W;L,K)=(X,AW' \cap KL;K,L) \stackrel{A}{=}(X,AW' \cap BC;B,C)$$ Hence $AW' \cap BC \equiv D$ proving the lemma.

Now let $AW$ intersect the incircle at $V',D'$ where $V'$ is closer to $A$ than $D'$.

Lemma 2: $\angle ADJ=\angle JDV'$

Proof: Let $AD$ intersect the incircle again at $Z$ from Lemma 1 we get $ZJ=JV'$ hence: $$\angle ADJ=\angle ZDJ=\widehat{ZJ}=\widehat{JV'}=\angle JDV'$$

Let $AJ$ intersect the perpendicular bisector of $MN$ at $U$

Lemma 3: $AMNDU$ is cyclic

Proof: $AMND$ is cyclic follows easily from $JE \parallel JM$ and $JF \parallel JN$ giving: $$\angle MAN=\angle EJN=180^{\circ}-\angle NDM$$ Also these parallel lines give $\angle NAJ=\angle FJI=\angle EJI=\angle MAJ$ so $AJ$ bisects $\angle MAN$ hence $U$ also lies on this circle

Lemma 4: $U$ lies on the incircle

Proof: From the previous observations: $$\angle DUJ=\angle DUA=\angle DMA=\angle DEJ$$ So $DUEJ$ are concyclic

Lemma 5: $\triangle ANM \simeq \triangle V'EF$ and they are oppositely orientated

Proof: Using Lemma 3 and then Lemma 2 and observing $DJ$ bisects $\angle FDE$ (as $J$ is arc midpoint) we get: $$\angle ANM=\angle ADM=\angle ZDE=\angle FDV'=\angle FEV'$$ And similarly $\angle NMA=\angle V'FE$

Lemma 6: $JV' \parallel MN$

Proof: Reflecting in line $DJ$ takes lines $DFN \leftrightarrow DEM$ and $DA \leftrightarrow DV'$ hence $\triangle ANM$ reflects to a triangle which is both directly similar and in perspective to $\triangle V'EF$ proving the claim

Lemma 7: $V'$ lies on the perpendicular bisector of $MN$

Proof: Notice that $\angle UV'J=90^{\circ}$ as $UJ$ is a diameter of the incircle hence as $U$ lies on the perpendicular bisector of $MN$ and $JV' \parallel MN$, $V'$ must also lie on this line.

Combining these results shows $V \equiv V'$ lies on the incircle as desired.

  • $\begingroup$ You get wrong points, but your idea is great! Thanks. $\endgroup$ – Quỳnh Vũ Thị Jan 15 at 2:24
  • $\begingroup$ Can you explain for me how to prove lemma 6? Thank $\endgroup$ – Quỳnh Vũ Thị Jan 16 at 14:45

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