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Source: China TST 1996, problem 1

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3647 posts

orl

#1 h May 17, 2005, 6:01 PM • 2 Y

Y by Adventure10, Mango247

Let side $BC$ of $\bigtriangleup ABC$ 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 $AM \perp BC$ .

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6555 posts

#2 h May 17, 2005, 6:16 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

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

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190 posts

who

#3 h Jun 19, 2006, 10:37 AM • 2 Y

Y by Adventure10, Mango247

,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.

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858 posts

Ashegh

#4 h Jun 27, 2006, 9:08 AM • 2 Y

Y by Adventure10, Mango247

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…

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1886 posts

Potla

#5 h Mar 15, 2011, 5:12 PM • 2 Y

Y by Adventure10, Mango247

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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1963 posts

#6 h Mar 15, 2011, 6:00 PM • 2 Y

Y by Adventure10, Mango247

$\square BCED$ 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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9782 posts

jayme

#7 h Mar 16, 2011, 8:40 AM • 2 Y

Y by Adventure10, Mango247

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

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