Let side BC of △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⊥BC.
Problem
Source: China TST 1996, problem 1
Tags: geometry, trapezoid, China
17.05.2005 21:16
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
19.06.2006 13:37
,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.
27.06.2006 12:08
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…
15.03.2011 19:12
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 △ABC. Now join AH,HM. Let K=BE∩DF,L=CD∩EG. Note that ∠BDF=π2−∠DBC=∠LCG, and ∠DFB=∠LGC=π2. Then △BDF∼△CGL. this leads to GLGC=BFDF. Analogously we have △KBF∼△CEG, implying KFFB=CGGE. These two jointly imply that GC⋅BF=GL⋅DF=KF⋅GE; so that we have EGDF=GLKF=EG−GLDF−KF=ELDK. On the other hand, note that M=FE∩DG⟹△DMF∼△MEG, so that EGDF=DMMG. But, △DHK∼△HEL⟹ELDK=DHHL. Using the last three relations, we note that DMMG=DHHL, which, from Thale's theorem, gives HM∥EG. Since EG⊥BC, we note that HM⊥BC, ie M lies on AH. So AM is the A-altitude of △ABC.◻
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15.03.2011 20:00
◻BCED is cyclic, DE∩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 , ¯AHM is the polar of I and AM⊥BC
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16.03.2011 10:40
Dear Mathlinkers, why not with Pappus theorem by considering the tangent to the circle at B and C... Sincerely Jean-Louis