Let $ABC$ be a triangle. Points $D$ and $E$ lie on sides $BC$ and $CA$, respectively, such that $BD=AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of angle $BCA$ meet segments $AD$ and $BE$ at $Q$ and $R$, respectively. Prove that $\frac{PQ}{AD}=\frac{PR}{BE}$.
Problem
Source: MOP 2005 Homework - Red Group #19
Tags: geometry, angle bisector, geometry unsolved
spikerboy
08.05.2014 06:01
Well the problem is just a bashing in disguise.$CR$ and $CQ$ are the $C$-angle bisector of the $\triangle ADC$ and $\triangle BEC$.So applying angle bisector theorem we get, \[\frac{QD}{QA}=\frac{CD}{AC} \implies \frac{AD}{QA}=\frac{CD+AC}{AC}=\frac{CD+CE+AE}{AC}.\] similarly we get,\[\frac{EC}{ER}=\frac{BD+EC+CD}{BE}\].Cancelling two common terms from these two equations we get, \[\frac{QA}{AC}\cdot \frac{EC}{ER}=\frac{AD}{BE}\] Applying menelaus on $\triangle APE$ with the transversal $CRQ$ we get, \[ \frac{ER}{PR}\cdot \frac{AC}{EC}\cdot \frac{PQ}{QA}= 1 \implies \frac{PQ}{AD}=\frac{PR}{BE}\]
jayme
23.05.2014 15:23
Dear Mathlinkers, revisiting this nice problem, a proof without calculation is possible... Any idea? Sincerely Jean-Louis
A.L.E.X
30.06.2021 15:23
Let's write down Angle Bisector equations.We have:
$$\frac{CE}{CB}=\frac{ER}{BR}\Leftrightarrow \frac{BR}{ER}=\frac{CD}{CE}+\frac{BD}{CE}\Leftrightarrow BD=\frac{BR\times CE-ER\times CD}{ER}$$$$\frac{CA}{CD}=\frac{AQ}{QD}\Leftrightarrow \frac{AE}{CD}=\frac{CE}{CD}=\frac{AQ}{QD}\Leftrightarrow AE=\frac{AQ\times CD-QD\times CE}{QD}$$And since $AE=BD$ we have:
$$AE=BD\Leftrightarrow \frac{AQ\times CD-QD\times CE}{QD}=\frac{BR\times CE-ER\times CD}{ER}\Leftrightarrow QD\times CE\times BE=CD\times ER\times AD$$Now using Menelaus Theorem in $\triangle BPD$ wrt $R,Q,C:$
$$\frac{CD\times PQ\times BR}{BC\times PR\times QD}=1\Leftrightarrow \frac{PQ}{PR}=\frac{BC\times QD}{CD\times BR}=\frac{ER\times AD\times BC}{CE\times EB\times BR}=\frac{ER\times AD\times CE}{CE\times EB\times ER}=\frac{AD}{EB}\Leftrightarrow \frac{PQ}{PR}=\frac{AD}{BE}$$