Points $D,E,F$ are midpoints of the sides $AB,BC,CA$ of triangle $ABC$. Angle bisectors of the angles $BDC$ and $ADC$ intersect the lines $BC$ and $AC$ respectively at the points $M$ and $N$, and the line $MN$ intersects the line $CD$ at the point $O$. Let the lines $EO$ and $FO$ intersect respectively the lines $AC$ and $BC$ at the points $P$ and $Q$. Prove that $CD=PQ$. (Plamen Koshlukov)
Problem
Source: Bulgaria 1992 P5
Tags: geometry, Triangles
hakN
18.05.2021 10:49
By the angle bisector theorem, we get $\frac{CM}{MB} = \frac{CD}{DB} = \frac{CD}{DA} = \frac{CN}{NA} \implies MN\parallel AB$. Let $EO \cap AB = H$ and $FO \cap AB = G$. Since $D$ is the midpoint of $AB$ and $MN\parallel AB$, we get that $O$ is the midpoint of $MN$. And since $GH\parallel EF$, we get that $D$ is also the midpoint of $GH$, implying that $AH = BG$. Now by Menelaus Theorem we get $\frac{QB}{QC} \cdot \frac{CF}{FA} \cdot \frac{AG}{GB} = 1 \implies \frac{QB}{QC} = \frac{GB}{GA}$. Also we get $\frac{PA}{PC} \cdot \frac{CE}{EB} \cdot \frac{BH}{HA} = 1 \implies \frac{PA}{PC} = \frac{HA}{HB}$. And since $HA = GB$ and $HB = GA$, we get $\frac{PA}{PC} = \frac{QB}{QC}$, so $PQ\parallel AB$. Now again by Menelaus, we get $\frac{PA}{PC} \cdot \frac{CO}{OD} \cdot \frac{DH}{HA} = 1$. Since $\frac{CO}{OD} = \frac{CM}{MB} = \frac{CD}{DB}$ and $\frac{DH}{HA} = \frac{DE}{PA}$, we get that $PC = \frac{CD \cdot AC}{AB}$. Finally, we have $\frac{AB}{PQ} = \frac{AC}{PC} = \frac{AB}{CD} \implies PQ = CD$, as desired.
zuss77
11.06.2021 15:33
Also as Spain 2017 P6