Orl, you could remember this one! [url=http://groups.yahoo.com/group/Hyacinthos/message/9531?expand=1]Hyacinthos message #9531[/url] wrote: From: "orl_ml" (Orlando Doehring) Subject: AM is perpendicular to BC Let side BC of triangle ABC be the diameter of a semicircle which cuts AB and AC at D and E respectively. F and G are the feet of the perpendiculars from D and E to BC respectively. DG and EF intersect at M. Prove that AM is perpendicular to BC. [url=http://groups.yahoo.com/group/Hyacinthos/message/9550?expand=1]Hyacinthos message #9550[/url] wrote: From: Darij Grinberg Subject: Re: AM is perpendicular to BC Dear Orlando, In Hyacinthos message #9531, you wrote: >> Let side BC of triangle ABC be the diameter >> of a semicircle which cuts AB and AC at D and >> E respectively. F and G are the feet of the >> perpendiculars from D and E to BC respectively. >> DG and EF intersect at M. Prove that AM is >> perpendicular to BC. Well, D and E are just the feet of the altitudes from C and B, respectively. Let S be the foot of the altitude from A, and let H be the orthocenter of triangle ABC. Let the line DG meet the line AS at M', and let the line EF meet the line AS at M". We will show that the points M' and M" coincide. Since the point M' lies on the lines AS and DG, the Menelaos theorem applied to triangle ABS yields AM' SG BD --- * -- * -- = -1, M'S GB DA and therefore AM' DA GB AD GB --- = - -- * -- = -- * --. M'S BD SG BD SG Since EG || HS, we have GB / SG = EB / HE, so that AM' AD EB AD DH EB AE --- = -- * -- = -- * -- * -- * --. M'S BD HE DH BD AE HE The triangles BDH and BEA are similar, so DH AE DH EB -- = --, and -- * -- = 1, BD EB BD AE so that our equation above simplifies to AM' AD AE AD AE --- = -- * -- = - -- * --. M'S DH HE DH EH Now, this expression is symmetric in B and C; hence, by analogy, AM" AD AE --- = - -- * --, M"S DH EH and the points M' and M" coincide. Thus, the lines DG, EF and AS meet at one point M' = M" = M, and AM is perpendicular to BC. There should be a simpler proof. Thanks for this and all the other problems, many of them are still too difficult or non-standard for me. Friendly, Darij Grinberg [url=http://groups.yahoo.com/group/Hyacinthos/message/9553?expand=1]Hyacinthos message #9553[/url] wrote: From: Vladimir Dubrovsky Subject: Re: [EMHL] Re: AM is perpendicular to BC Dear Darij, here is a shorter solution to Hyacinthos message #9531 -- > >> Let side BC of triangle ABC be the diameter > >> of a semicircle which cuts AB and AC at D and > >> E respectively. F and G are the feet of the > >> perpendiculars from D and E to BC respectively. > >> DG and EF intersect at M. Prove that AM is > >> perpendicular to BC. Let D', E' be reflections of D, E in BC, N the intersection of DE' and D'E. Obviously, N lies on BC and MN is parallel to DD' and EE', or perpendicular to BC (because M and N divide DG and DE', resp., in the same ratio DF:EG=DD':EE'). So it remains to show that, in your notation, N=S, the foot of the altitude from A, or, equivalently, that angle NEC = angle ABC.. But angle NEC = angle D'EC = angle DE'C = angle DBC. QED Best, Vladimir Dubrovsky Darij

,although it noway explians why result is true,it admits inelegent but short co ordinate sol,take A-0,1,B-b,oand C-c,o.

Suppse $DE$, intersects $BC$ at $S$.quadrilateral $DECB$ is perfect, then, the polar of $S$, to circle. Goes through $A$.suppose $DF,EG$ intersects the circle at points $D',E'$ respectively. $DD',EE'$ are parallel , then quadrilateral $DEE'D'$ is a trapezoid, and $D'E'$ also meet $BC$, at $S$. then $DEE'D'$, is also perfect, and the polar of $S$, to circle , goes through the intersection of $ED',DE'$. If we name this point, $K$, it is clear that it lies on $BC$. ............................................................................................................................................................................................. Then it is concluded that polar of $S$, to circle, is line $AK$, which is perpendicular to $BC$. $(1)$ ............................................................................................................................................................................................. Now we use papus theorem , for points $EGE'$ and $DFD'$, on the two parallel lines, According to this theorem we have: The intersection of $EF,DG=M$ And , the intersection of $FE',GD'=N$ And , the intersection of $ED',DE'=K$ Are collinear. Because of $DEE'D'$ is a trapezoid,and because of the symmetry of the shape. ............................................................................................................................................................................................. $M,N$, turn into each other with a respect to line $BC$, and it means that $MK$ is perpendicular to $BC$. $(2)$ ............................................................................................................................................................................................. Then according to $(1),(2)$: $AK$ is perpendicular to $BC$. And $MK$ is also perpendicular to $BC$. And it means that, $A,M,K$ are collinear, and $AM$ is perpendicular to $BC$. And every thing is ok. I didn’t enjoy my solution…

I have got another solution today, so I would like to post it here. Sorry for reviving this old topic, but it is always better not to create a new topic! Obviously $CD$ and $BE$ are the altitudes of the triangle $ABC,$ so let them intersect at the orthocentre $H$ of $\triangle ABC.$ Now join $AH,HM.$ Let $K=BE\cap DF, L=CD\cap EG.$ Note that $\angle BDF=\frac{\pi}{2}-\angle DBC=\angle LCG,$ and $\angle DFB=\angle LGC=\frac{\pi}{2}.$ Then $\triangle BDF\sim \triangle CGL.$ this leads to $\frac{GL}{GC}=\frac{BF}{DF}.$ Analogously we have $\triangle KBF\sim\triangle CEG,$ implying $\frac{KF}{FB}=\frac{CG}{GE}.$ These two jointly imply that $GC\cdot BF=GL\cdot DF=KF\cdot GE;$ so that we have $\frac{EG}{DF}=\frac{GL}{KF}=\frac{EG-GL}{DF-KF}=\frac{EL}{DK}.$ On the other hand, note that $M=FE\cap DG\implies \triangle DMF\sim\triangle MEG,$ so that $\boxed{\frac{EG}{DF}=\frac{DM}{MG}}.$ But, $\triangle DHK\sim\triangle HEL\implies \boxed{\frac{EL}{DK}=\frac{DH}{HL}}.$ Using the last three relations, we note that $\frac{DM}{MG}=\frac{DH}{HL},$ which, from Thale's theorem, gives $HM\parallel EG.$ Since $EG\perp BC,$ we note that $HM\perp BC,$ ie $M$ lies on $AH.$ So $AM$ is the $A$-altitude of $\triangle ABC.\Box$

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$\square BCED$ is cyclic, $DE\cap BC=I$ and $A$ is on the polar of $I$ (Brocard) Draw a line through $M$ perpendicular to $BC$ and let it cut $DE$, $BC$ at $H$, $J$ $FJ/JG=DF/EG=FI/GI$, that is, ($F$,$G/J$,$I$)$=-1=$ ($D$,$E/H$,$I$) Therefore, $HM$ is the polar of $I$ , $\overline{AHM}$ is the polar of $I$ and $AM\bot BC$

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Dear Mathlinkers, why not with Pappus theorem by considering the tangent to the circle at B and C... Sincerely Jean-Louis