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Romanian Mathematical Olympiad - Final Round

Geometry Difficulty 5.0 AIME Prove it Romania

In triangle $ABC$ , the bisector $AD$ and the median $BE$ meet at $P$ . The straight lines $AB$ and $CP$ meet at $F$ . The parallel through $B$ at $CF$ meets the straight line $DF$ at $M$ . Show that $DM = BF$ .

Gheorghe Bumbăcea

Solution

Ceva's Theorem implies
$\frac{BF}{FA} \cdot \frac{AE}{EC} \cdot \frac{CD}{DB} = 1,$
hence $\frac{BF}{FA} = \frac{BD}{DC}$ . The converse of Thales' Theorem yields $DF \parallel AC$ . From $\triangle BFD \sim \triangle BAC$ follows $\frac{BF}{BA} = \frac{FD}{AC}$ , whence $BF = \frac{AB}{AC} \cdot FD$ . Now $\triangle BDM \sim \triangle CDF$ implies $\frac{DM}{FD} = \frac{BD}{DC}$ . On the other hand, $BD$ being a bisector, $\frac{BD}{DC} = \frac{AB}{AC}$ , so $DM = \frac{AB}{AC} \cdot FD$ . Finally, $DM = BF$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.