## Prove concyclic

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Tahmid Hasan
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### Prove concyclic

Let $ABCD$ be a cyclic quadrilateral. Let $AD \cap BC=E,AC \cap BD=F,EF \cap CD=G$. Let $M$ be the midpoint of $CD$. Prove that $A,B,M,G$ are con-cyclic.
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### Re: Prove concyclic

Let $AB \cap DC =H$

Note that $D,G,C,H$ build a harmonic range .
So

$\displaystyle \frac{DG}{CG}=\frac{DH}{CH}$
$\Rightarrow MC^2=(DG-MG)(CG+MG)=MG.MH$
$\Rightarrow DG.CG=MG.GH$

Now $HB.HA=HC.HD=HG.HM$
$\Leftrightarrow HC.(HC+CD)=(HC+CG).(HC+CM)$
$\Leftrightarrow HC.CD=HC.CM+HC.CG+CG.CD$
$\Leftrightarrow HC.MG=CG.CD$
$\Leftrightarrow \displaystyle \frac{HC}{CG}=\frac{CM}{MG}=\frac{DH}{DG}$
$\Leftrightarrow \displaystyle \frac{CG}{MG}=\frac{GH}{DG}$
Last edited by Nadim Ul Abrar on Wed Feb 27, 2013 1:46 pm, edited 1 time in total.
$\frac{1}{0}$

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Prove concyclic

Nadim Ul Abrar wrote:Let $AB \cap DC =H$

Note that $D,G,C,H$ build a harmonic range .
So

$\displaystyle \frac{DM}{CM}=\frac{DH}{CH}$
I don't think so.
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### Re: Prove concyclic

Nadim Ul Abrar wrote: Note that $D,G,C,H$ build a harmonic range .
So
$\displaystyle \frac{DM}{CM}=\frac{DH}{CH}$
I think that's a typo, and hope Nadim vai would post a correction.

My solution:

Let $AB \cap DC =$ $H$, $O$ the center of $(ABCD)$, $OH \cap EF = I$, $r=$ circumradius of $(O)$.
Now,, as $\angle OMG = \angle OIG = 90^\circ$, $IOMG$ concyclic.
Again note that, $I$ is the inverse of $H$ wrt $(O)$ (as $EF =$ polar of $H$ wrt $(O)$ ).

Now, $MH.HG = OH.HI = OH^2 - OI.OH = OH^2 - r^2 = AH.HG$

So, $A,B,M,G$ are concyclic.

Use $L^AT_EX$, It makes our work a lot easier!

Tahmid Hasan
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Joined: Thu Dec 09, 2010 5:34 pm
My solution is pretty computational. At first I would like to note that in my figure $CD \cap BA=H$.
From the property of harmonic division we get $\frac{DH}{CH}=\frac{DG}{CG}=\frac{DM-GM}{DM+GM}$
$\Rightarrow DH.DM+DH.GM=CH.DM-CH.GM \Rightarrow DM.DM+DM.GM=(DH+2DM)DM-(DH+2DM)GM$
$\Rightarrow DM^2=GM.DH+DM.GM$.
$A,B,M,G$ are concyclic $\Leftrightarrow HA.HB=HG.HM \Leftrightarrow HD.HC=HG.HM$
$\Leftrightarrow DH(DH+2DM)=(DH+DM-GM)(DH+DM) \Leftrightarrow DM^2=GM.DH+DM.GM$.