Let $P$ be the intersection point between $AN$ and $CM$, and let $AN$ intersect $CD$ in $Q$ (I supposed that $AB<BC$, otherwise we switch letters, but it's the same proof). Then $DP$ is the interior angle bisector if and only if (the bisector theorem) \[ \frac{ AP}{PQ} = \frac{AD}{DQ} \quad (1) \] But the triangles $AMP$ and $QPC$ are similar, thus \[ \frac{AP}{PQ} = \frac{AM}{CQ} = \frac{CN} {CQ } = \frac{AD}{DQ} \] where the last equality is derived from Thales for the triangle $QAD$ and the parallel $CN$.

You have one more...You are sharp I wonder : Do you Valentin would be agree to participate to IMO as a french candidate ? Pierre.

I can't, I'm over the age limit , and when I had the limit I represented Romania

Pfffff.....really too bad.... Pierre.

Valentin Vornicu wrote: (good thing Darij isn't awake now, so I can post first :P) Actually, I was awake but I am currently not too active on the forum since I'm at the Oberwolfach IMO training seminar. (We are actually forbidden to use Internet at all, but nobody cares.) The problem was posed on the first round of the BWM 2003. I had solved it in December 2002, but to that time I was not yet "the" synthetic geometer I am now, and I solved it using the Sine Law... so, Valentin, your solution is better. See also the synthetic solutions at the official Bundeswettbewerb website, http://bundeswettbewerb-mathematik.de/aufgaben/aufgaben.htm#Loesungen Loesungen 2003.1 (in German). Darij

i v got the sam sol that posted valentin

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pbornsztein wrote: Alternative formulation. Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$. We can always use a homographic transformation : The map $M\mapsto N$ is homographic between the lines $AB,CB$ (with points $A,C$ considered as origins on these lines respectively). The locus of the point $Q$ is then determined by any three points on this locus. We can choose the cases when $Q$ lies on $CB,AB$ and the line at infinity. These cases are solved easily and we find that $Q$ lies on the desired bisector. This is totally overkill, so I'll definitely think about it a bit longer .

the easist is mine: only use the law of sines for 4 triangels

Here is my proof: Let $P$ be the intersection of $PQ$ and $AC$. Then: $\frac{AP}{CP}=\frac{\triangle{ADQ}}{\triangle{CDQ}}=\frac{\triangle{ADQ}}{\triangle{CNQ}}\cdot\frac{\triangle{CNQ}}{\triangle{AMQ}}\cdot\frac{\triangle{AMQ}}{\triangle{CDQ}}$ $=\frac{AD\cdot AQ}{CN \cdot QN}\cdot \frac{CQ \cdot QN}{AQ\cdot MQ}\cdot \frac{AM\cdot QM}{CQ \cdot MQ}$ $=\frac{AD}{CD}$. Cheers!

Let $X=MC\cap AD$. Notice that $\triangle XAM\sim\triangle XDC$, $\triangle XAQ\sim\triangle CNQ$. So \[ \frac{XQ}{QC}=\frac{XA}{CN}=\frac{XA}{AM}=\frac{XD}{DC} \]Now done by angle bisector theorem.

Let $P=DQ\cap AB$ we show $AP=AD$ or since $AP=AM+MP$ and $AD=BC=BN+NC$ we only need to prove $MP=BN$. By Menelaus on $\triangle MBC$ (Points $A,Q$ and $N$ are collinear) we have $$\frac{QC}{MQ}= \frac{AB}{NB} $$Note that $\triangle QMP\sim \triangle QCD$ so $\frac{QC}{MQ} = \frac{DC}{MP} $ Therefore $MP=BN$

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