ED is perpendicular to FD (CGMO 2010-P2)

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ED is perpendicular to FD (CGMO 2010-P2)

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#1 h Aug 17, 2010, 3:39 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

In triangle $ABC$ , $AB = AC$ . Point $D$ is the midpoint of side $BC$ . Point $E$ lies outside the triangle $ABC$ such that $CE \perp AB$ and $BE = BD$ . Let $M$ be the midpoint of segment $BE$ . Point $F$ lies on the minor arc $\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \perp BE$ . Prove that $ED \perp FD.$

$[asy] defaultpen(fontsize(10)); size(6cm); pair A = (3,10), B = (0,0), C = (6,0), D = (3,0), E = intersectionpoints( Circle(B, 3), C--(C+100*dir(B--A)*dir(90)) )[1], M = midpoint(B--E), F = intersectionpoints(M--(M+50*dir(E--B)*dir(90)), circumcircle(A,B,D))[0]; dot(A^^B^^C^^D^^E^^M^^F); draw(B--C--A--B--E--D--F--M^^circumcircle(A,B,D)); pair point = extension(M,F,A,D); pair[] p={A,B,C,D,E,F,M}; string s = "A,B,C,D,E,F,M"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} d[4] = -50; string[] k= split(s,","); for(int i = 0;i<p.length;++i) { label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i])); }[/asy]$

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#2 h Aug 17, 2010, 6:54 PM • 2 Y

Y by Adventure10, Mango247

You may see here! http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=361789&hilit=CGMO+2010+p2

Best regards,
sunken rock

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SKhan

#3 h Aug 18, 2010, 10:42 AM • 4 Y

Y by ArseneLupin, Adventure10, Mango247, Artist_of_Math

I have just copied and pasted my post.

SKhan wrote:

Here is my solution for problem 2.

Let the perpendicular bisector of $BE$ and the perpendicular of $DE$ at $D$ meet at the point $P$ .
$\angle PDE=90^o=\angle PME$ so $PDME$ is a cyclic quadrilateral.
Note that $\triangle BPE$ is isosceles so $\angle EPM=\angle MPB$ .
$D$ and $M$ are the midpoints of $BC$ and $BE$ so $DM\parallel CE$ so $\angle MDB=\angle ECB$
$\angle BPD=\angle MPD-\angle MPB=\angle MED-\angle EPM=\angle BED-\angle EDM$
$\angle BPD=\angle EDB-\angle EDM=\angle MDB=\angle ECB=90^o-\angle DBA=\angle BAD$
Hence $ABDP$ is a cyclic quadrilateral.
$P$ and $F$ are on the same side of the diameter $AB$ so $P$ and $F$ coincide.
So $ED\perp FD$ as required.

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#4 h May 8, 2013, 2:47 AM • 3 Y

Y by Sx763_, Richangles, Adventure10

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#5 h Nov 19, 2017, 3:13 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A=(0,120), B=(-40,0), C=(40,0), D=(0,0), E=(-61.4,33.8), F=(41.4,75.2), G=(-18.6,-33.8), H=(-82.8,67.6), M=(-50.7,16.9); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(M); draw(A--B--C--cycle, linewidth(0.5)); draw(circumcircle(A,D,B), linewidth(0.5)); draw(B--E--D, linewidth(0.5)); draw(D--H--E, linewidth(0.4)+dashed); draw(B--G--D, linewidth(0.4)+dashed); draw(D--F--M, linewidth(0.5)); label("$A$", A, (0,1)); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, (0,-1)); label("$E$", E, SW); label("$F'$", F, (1,0)); label("$G$", G, SW); label("$H$", H, (-1,0)); label("$M$", M, SW); draw(C--E, linewidth(0.4)+dashed); draw(H--F, linewidth(0.4)+dashed); [/asy]$
Define $F'$ as the point on the perpendicular bisector of $\overline{BE}$ satisfying $\overline{DE}\perp\overline{DF'}$ , $G$ as the intersection of $\overline{DF'}$ and $\overline{BE}$ , and $H$ as the second intersection of $\overline{BE}$ with $(ABD)$ . To begin, we must have that
$\angle BHD=90^\circ-\angle ABD=\angle BCE,$ which used in conjunction with $BD=BE$ implies that $\triangle BCE\cong\triangle BHD$ ; therefore, $EH=BE$ . Furthermore, since $\overline{BD}$ is the median to hypotenuse of $EDG$ , it follows that $GB=BE$ as well. Combining our length results with the fact that $EMDF'$ is cyclic yields
$GB\cdot GH=GM\cdot GE=GD\cdot GF',$ which completes our problem by POP.

This post has been edited 6 times. Last edited by Blast_S1, Jun 10, 2018, 8:59 PM

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#6 h May 2, 2020, 5:10 AM

Y by

Let $MF$ meet $AD$ at $X$ , let $N$ be the midpoint of $AB$ , now notice that $X$ lies on the perpendicular bisector of $EB$ and $BC$ so it is the circumcenter of $ABC$ .
Since $M,B,D,X$ are concyclic,
$\angle MXB=\angle MDB=\angle ECB=\angle BAD$ Hence $\triangle EXB\sim\triangle BAC$
Moreover, notice that $\angle EBA=\angle EBX-\angle ABX=\angle ABD-\angle ABX=\angle XBD$ , and
$\frac{BX}{AB}=\frac{EB}{BC}=\frac{1}{2}$ therefore, $BN=BX$ , so B lies on the perpendicular bisector of $NX$ and $ED$ which implies that $NXDE$ is an isoceles trapezoid.
We further have
$EX=XB=NB=NF$ and $\angle NEX=\angle NDX=\angle DAN=\angle DAB=\angle EXM$ So $EN\|MX$ , combined with the fact that $E$ and $F$ lies on different side of $NX$ , we conclude that $ENFX$ is a parallelogram. Hence $XD=EN=XF$
Now $\angle DXF=\angle MBD$ , hence $D$ is the center of spiral similarity sending $EB$ to $XF$ , hence also the center of spiral similarity sending $BX$ to $EF$ , so $\angle EDF=\angle BDX=90^{\circ}$
This completes the proof.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#8 h Jan 27, 2022, 4:08 PM

Y by

Let S be where perpendicular at M to BE and perpendicular at D to DE meet. we will show S is F. Clearly SDME is cyclic so ∠MSD = ∠MED = ∠BDE and
∠BSM = ∠MSE = ∠MDE. ∠DSB = ∠DSM - ∠BSM = ∠BDE - ∠MDE = ∠BDM = ∠DAB so DSAB is cyclic and S is F.
we're Done.

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