Let $D$ be an arbitrary point on side $AB$ of triangle $ABC$. Circumcircles of triangles $BCD$ and $ACD$ intersect sides $AC$ and $BC$ at points $E$ and $F$, respectively. Perpendicular bisector of $EF$ cuts $AB$ at point $M$, and line perpendicular to $AB$ at $D$ at point $N$. Lines $AB$ and $EF$ intersect at point $T$, and the second point of intersection of circumcircle of triangle $CMD$ and line $TC$ is $U$. Prove that $NC=NU$
Problem
Source: Bosnia and Herzegovina TST 2015 day 1 problem 2
Tags: geometry, circumcircle, perpendicular bisector
Luis González
16.05.2015 20:55
Since $\angle ADE=\angle ACB=\angle BDF$ $\Longrightarrow$ $AB$ bisects $\angle EDF$ externally $\Longrightarrow$ $ND$ bisects $\angle EDF$ internally $\Longrightarrow$ $M,N$ are midpoints of the arcs $EDF$ and $EF$ of $\odot(DEF)$ $\Longrightarrow$ $NE=NF$ and $\angle ENF=180^{\circ}-\angle EDF=180^{\circ}-(180^{\circ}-2\angle ACB)=2\angle ACB$ $\Longrightarrow$ $N$ is circumcenter of $\triangle CEF.$ Now, since $TE \cdot TF=TD \cdot TM=TU \cdot TC$ $\Longrightarrow$ $U \in \odot(CEF)$ or $NC=NU.$
samithayohan
05.06.2015 20:46
Its realy "Lots and lots of circles". My obvious fist attempt was inversion ,without any succes.So here goes an elementary proof by angle chasing. First we are going to show $FDME$ are cyclic.Assume that $M'$ is the second intersection of $\odot$$FDE$ with$AB$. $\angle$$M'EF$$=$$180$-$\angle$$M'FE$-$\angle$$EM'F$$=$$180$-$\angle$$FDE$-$\angle$$M'DE$($FDM'E$ cyclic)$=$$180$-($180$-$\angle$$M'DE$-$\angle$$ADE$)-$\angle$$M'DE$. $ADFC$ cyclic and $BDEC$ cyclic $\implies$ $\angle$$ADE$$=$$\angle$$BDF$$=$$\angle$$ECF$ Hence$\angle$$M'EF$$=$$\angle$$ECF$.On the other hand we have that $\angle$$M'FE$$=$$\angle$$ECF$ $\implies$ $\angle$$M'EF$$=$$\angle$$M'FE$ $\implies$ $M'E$$=$$M'F$ Hence $M'$ must lie on the perpendicular bisectosr of $EF$ $\implies$ $M'$$=$$M$ $\implies$$EMDF$ cyclic. Another easy angle chasing will tell that $N$ lies on $\odot$$EMDF$ The rest of my proof is very similar to Luis Gonzalezs solution. Quote: $\angle ENF=180^{\circ}-\angle EDF=180^{\circ}-(180^{\circ}-2\angle ACB)=2\angle ACB$ $\Longrightarrow$ $N$ is circumcenter of $\triangle CEF.$ Now, since $TE \cdot TF=TD \cdot TM=TU \cdot TC$ $\Longrightarrow$ $U \in \odot(CEF)$ or $NC=NU.$ . Hence we are done.