In the triangle $ABC$, the angle - bisector $AD$ ($D \in BC$) and the median $BE$ ($E \in AC$) intersect at point $P$. Lines $AB$ and $CP$ intesect at point $F$. The parallel through $B$ to $CF$ intersects $DF$ at point $M$. Prove that $DM = BF$
Problem
Source: 2013 Romanian National MO grade 7 P1
Tags: geometry, angle bisector, median, parallel
jlordhe
25.07.2018 02:00
Romania? Is that next to Russia or something like that?
parmenides51
25.07.2018 02:02
Romania, it's a country in the Balkan (Europe)
jayme
30.07.2018 15:29
Dear Mathlinkers, any ideas? Sincerely Jean-Louis
sunken rock
30.07.2018 15:59
Since $BE$ is median, $FD\parallel AC$, thus $\frac{BF}{AF}=\frac{BD}{CD}\ (\ 1\ )$. From angle bisector theorem, $\frac{BD}{CD}=\frac{AB}{AC}\ (\ 2\ )$ and $FD=AF\ (\ 3\ )$. From $(1)\wedge (2)$ we get $\frac{BF}{AF}=\frac{AB}{AC}\ (\ 4\ )$. $\triangle BMF\sim\triangle FCA\implies\frac{BF}{AF}=\frac{FM}{AC}\ (\ 5\ )$. From $(4)\wedge (5)$ we get $FM=AB\ (\ 6\ )$; with $(3)$ we are done. Best regards, sunken rock
jayme
02.08.2018 11:56
Dear Mathlinkers, Thank S.F. four your proof... see : https://artofproblemsolving.com/community/c6h1683651_two_parallel which help... Sincerely Jean-Louis