Let $ABC$ be a triangle and $AA_1$ be the angle bisector of $A$ ($A_1 \in BC$). The point $P$ is on the segment $AA_1$ and $M$ is the midpoint of the side $BC$. The point $Q$ is on the line connecting $P$ and $M$ such that $M$ is the midpoint of $PQ$. Define $D$ and $E$ as the intersections of $BQ$, $AC$, and $CQ$, $AB$. Prove that $CD=BE$.
Problem
Source: Bulgaria JBMO TST 2018, Day 2, Problem 2
Tags: geometry, angle bisector
Devastator
25.06.2018 11:26
Will all these JBMO TST problems get complied and placed in the contest collections? (I am replying here as to not flood this topic too much) Cool, will try them eventually.
Amir Hossein
25.06.2018 11:27
Yeah, that's my intention for Bulgaria JBMO TST 2018. EDIT. Done.
Devastator
25.06.2018 11:57
We can see that BPCQ is a parallelogram
So I noticed that $BE=CD$ iff the circumradii of triangles BQE and CQD are equal, so maybe it can be proven by using the other sides and angles and trig
Edit: Completed it
Since BM=CM and PM=QM, then BPCM is a parallelogram
Label the angles AEC, and ADB as $x,y$ respectively.
Notice that by Extended Sine Law, $BE=CD$ iff
$$circumradius(BQE)=\frac{BE}{sinBQE}=\frac{CD}{sinCQD}=circumradius(CQD)$$Hence, it suffices to show that $\frac{BQ}{CQ}=\frac{sin(x)}{sin(y)}$
But notice that by Sine Law and the fact that BP parallel to CE, CP parallel to BD,
$$\frac{BQ}{CQ}=\frac{CP}{BP}=\frac{ \frac{CP}{sin(CAP)} }{ \frac{BP}{sin(BAP)} }=\frac{ \frac{AP}{sin(ABP)}{ \frac{AP}{sin(ACP)} }=\frac{sin(x)}{sin(y)}$$
Can someone pls help me find my latex error?
dionadam
25.06.2018 11:57
Using Law of Sinus we deduce that $\dfrac{BE}{BC}=\dfrac{\sin{\widehat{BCE}}}{\sin{\widehat{BEC}}}$ $\dfrac{CD}{BC}=\dfrac{\sin{\widehat{CBD}}}{\sin{\widehat{CDB}}}$ As a result we conclude that it suffices to prove that $\dfrac{\sin{\widehat{BCE}}}{\sin{\widehat{BEC}}}=\dfrac{\sin{\widehat{CBD}}}{\sin{\widehat{CDB}}}$ Its is easy to see that $BPCQ$ is parallelogram ($M$ is the midpoint of both $BC$ and $PQ$). Consequently $\widehat{BCE}=\widehat{CBP}, \widehat{BEC}=\widehat{ABP}, \widehat{CBD}=\widehat{BCP}, \widehat{CDB}=\widehat{ACP}$. Therefore it suffices to prove that $\dfrac{\sin{\widehat{CBP}}}{\sin{\widehat{ABP}}}=\dfrac{\sin{\widehat{BCP}}}{\sin{\widehat{ACP}}}$ Applying again Law of Sinus we deduce that $\dfrac{\sin{\widehat{CBP}}}{\sin{\widehat{BPA_1}}}=\dfrac{\sin{\widehat{A_1BP}}}{\sin{\widehat{BPA_1}}}=\dfrac{PA_1}{BA_1}$ $\dfrac{\sin{\widehat{ABP}}}{\sin{\widehat{BPA}}}=\dfrac{PA}{BA}$ However, since $\widehat{BPA_1}+\widehat{BPA}=180^o$, it is $\sin{\widehat{BPA_1}}=\sin{\widehat{BPA}}$. As a result $\dfrac{\sin{\widehat{CBP}}}{\sin{\widehat{ABP}}}=\dfrac{PA_1\cdot BA}{BA_1\cdot PA}$. For the same reasons, $\dfrac{\sin{\widehat{BCP}}}{\sin{\widehat{ACP}}}=\dfrac{PA_1\cdot CA}{CA_1\cdot PA}$ It suffices to prove that $\dfrac{BA}{BA_1}=\dfrac{CA}{CA_1}\Leftrightarrow \dfrac{BA}{CA}=\dfrac{BA_1}{CA_1}$, which is true because of the wellknow "bisector theorem".
PROF65
25.06.2018 18:22
Let $L=BP\cap AC,K=CP\cap AB$ , remark that $QD \parallel BL,QE \parallel CK$ since $PQ,BC$ have the same midpoint thus $\frac{KB}{KA}=\frac{EC}{CA},\frac{LC}{LA}=\frac{DB}{BA}$ but by Ceva's we deduces $\frac{KB}{KA}.\frac{LA}{LC}=\frac{BA_1}{A_1C}=\frac{AB}{AC}\implies \frac{EC}{CA}.\frac{BA}{BD}=\frac{AB}{AC}$ therefore the result follows . RH HAS
AlastorMoody
26.05.2019 05:05
This is just Sine Rule on $\Delta APB $, $\Delta APC $, $\Delta BEQ $ and $\Delta CQD $