mr.danh wrote: ... PM meets QN at I... :

Jutaro wrote: Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA = M$ and $ AQ \cap BC = N$, show that $ MN$, $ r$ and $ AC$ concur. - $ AC\cap r = D$, sufficient to show that: $ M,N,D$ are colinear. - $ AN\cap CM = U; BU\cap AC = E$. Because: $ \noindent \angle{PBA} = \angle{QBC}\Rightarrow \triangle{PBA}\sim\triangle{QBC}\Rightarrow \dfrac{QB}{PB} = \dfrac{QC}{PA} = \dfrac{QU}{AU}\Rightarrow EB\perp PQ\Rightarrow \angle{ABE} = \angle{CBE}.$ Then, because $ AN,CM,BE$ are concurrent, $ N,M,D$ are colinear if and only if $ (AECD)$ forms an harmonic division (lemma 1 http://reflections.awesomemath.org/2007_4/harmonic_division.pdf) and it's trivial (lemma 2 http://reflections.awesomemath.org/2007_4/harmonic_division.pdf).

suppose without loss of generality that $ C > A$.. and denote by $ T$ the intersection of $ AC$ and $ r$... we have that $ C$ lies between $ A$ and $ T$... we can prove that $ \dfrac{\sin BTN}{\sin CTN} = \dfrac{\sin BTM}{\sin CTM} = \dfrac{1}{2\cos\frac {A - C}{2}}$, and we're done... *this is not immediate, i mean, it requires a trig ceva and some law of sines, but it's easy..

I cant speak and write english The Desargues solution is the most popular We have to prove that APM and CQN are in perspective, and we know that AP//CQ, therefore we have to prove that if X is the intersection of PC and AQ BX is the angle bisector of <ABC o BX//CQ//AP. Let Y the intersection of CQ and AB. APB y CQB are similar then: AB/BC=AP/CQ APX y CQX too: AB/BC=AX/QX But CBY is isosceles and then BC=BY, AB/BC=AB/BY=AX/QX y BX//CQ.

Concyclicboy wrote: The solution based on the Desargues theorem, is the most popular. We have to prove that $ \bigtriangleup APM$ and $ \bigtriangleup CQN$ are perspective, and we know that $ AP\parallel CQ,$ therefore we have to prove that $ BX\parallel CQ\parallel AP,$ where $ X\equiv PC\cap AQ.$ Let $ Y$ be, the intersection of $ CQ$ and $ AB.$ $ \bigtriangleup APB,\ \bigtriangleup CQB$ are similar and then: $ \frac {AB}{BC} = \frac {AP}{CQ}.$ Also, $ \bigtriangleup APX,\ \bigtriangleup CQX$ are similar and then: $ \frac {AB}{BC} = \frac {AX}{QX}$ But $ \bigtriangleup CBY$ is isosceles and then $ BC = BY,$ and $ \frac {AB}{BC} = \frac {AB}{BY} = \frac {AX}{QX}$ $ \Longrightarrow$ $ BX\parallel CQ\parallel AP.$ I like this solution dear conyclicboy and let me to try a bit simplification. From the similar right triangles $\bigtriangleup PAB,\ \bigtriangleup QCB,$ we have also $\frac {AP}{CQ} = \frac {BP}{BQ},\ (1)$ From $AP\parallel CQ,$ by Thales theorem, we have that $\frac {AP}{CQ} = \frac {XP}{XC},\ (2)$ From $ (1),\ (2),\ \Longrightarrow$ $\frac {BP}{BQ} = \frac {XP}{XC}\ \ ,(3)\ \Longrightarrow$ $ BX\parallel CQ\parallel AP.$ Best regards, Kostas vittas.

Why does the intersection of PC with QA being in the angle bisector imply the triangles are in perspective?

oh wait oops... for a second there I dont know why i read concyclicboy was trying to prove that PXA and QXC were in perspective...

Concyclicboy wrote: We have to prove that $ \bigtriangleup APM$ and $ \bigtriangleup CQN$ are perspective, and we know that $ AP\parallel CQ,$ therefore we have to prove that $ BX\parallel CQ\parallel AP,$ where $ X\equiv PC\cap AQ.$ I was very confused with this. I think instead we have to show $ \bigtriangleup ABC$ and $ \bigtriangleup PXQ$ in perspective. Thus when $BX \parallel AP \parallel CQ$ is proved, descargues theorem shows that due to $AB \cap PX = M$, $AC \cap PQ = D$ and $BC \cap XQ =N$, $MDN$ is collinear.

Dear Mathlinkers, an alternative synthetic proof: 1. X the point of intersection of AP and BC, Y of CQ and BC 2. by Pappus theorem applied to the hexagon PAMNCQP inscribed in the lines AQ and CR, XY, PQ and MN are concurrent . 3. P is the midpoint of AX, Q the midpoint of CY 4. by considering the complete trapez XACY, we are done. Sincerely Jean-Louis

Let $S = r \cap AC $, then, let $D$ be the intersection of the angle bisector of $ \angle ABC $ and $AC$, then is well known that $BD$ is parallel to $AP$ and $CQ$ (since the three are perpendicular to $r$) Then, the problem can be read as "Prove that $AB$, $CP$ and $SN$ concur at $M$" Now let's suppose that $N'=SM \cap BC$, we want to prove $N'=N$ By Ceva Theorem, we have that $\frac{SA}{AC} \cdot \frac{CN'}{N'B} \cdot \frac{BP}{PS} = 1$ Then, by Menelaus Theorem with $\triangle SBC$ and points $A$, $N$ and $Q$ we have that $\frac{SA}{AC} \cdot \frac{CN}{NB} \cdot \frac{BQ}{QS} = 1$, then, we have $ \frac{CN'}{N'B} \cdot \frac{BP}{PS} = \frac{CN}{NB} \cdot \frac{BQ}{QS} $ so, it's clear that if we show that $\frac{BP}{PS} =\frac{BQ}{QS} $ then we are done, but, that would mean that $S$, $P$, $B$ and $Q$ are on harmonic division, but that is true; let $O$ be the "ideal point" (the infinite one) Then it is clear that this happens: $SADC\frac{O}{\barwedge} SPBQ$, but, $S$, $A$, $D$ and $C$ is a well known harmonic division (Internal and external angle bisector), so we are done