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This is a question that was asked on the site a few days back and before an answer, it was deleted by the OP. While I was able to solve the problem, I feel there is a more elegant way.

Question:

In $\triangle ABC$, $BF \perp AC$ and $AG \perp BC$. Also $E$ is the midpoint of $AB$.

Show that the circumcircle of $\triangle EFG$ passes through the circumcenter of $\triangle CFG$.

enter image description here

My solution:

Let $N$ be the intersection of the perpendicular bisector of $FG$ and the circumcircle of $\triangle EFG$. I assumed circumradius of $\triangle ABC = R$. Then I showed circumradius of $\triangle EFG = \frac{R}{2}$. Using similarity of $\triangle CGF$ and $\triangle CAB$, I showed $MN = k^2R$ where $FG:AB = CF:BC = CG:AC = k:1$ so circumradius of $\triangle CFG = kR$. Finally I showed $FN = NG = kR$ which proves $N$ is the circumcenter of $\triangle CFG$.

Would like to see other solutions.

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enter image description here

$\angle HFC=90^{\circ}=\angle HGC$. So $CFHG$ is cyclic. Circumcircle of $CFG$ is the same as circumcircle of quadrilateral $CFHG$. $CH$ is its diameter. Midpoint of $CH$, $N$, is its center.

Now $FE$ is the median of right $\triangle AFB$ and $GE$ is median of right $\triangle AGB$. So $AE=FE=GE=BE=a/2$.

In isosceles $\triangle BEG$, $\angle EGB=B$. Since $CH$ is part of altitude on $AB$, $\angle NCG=90-B$. In isosceles $\triangle CNG$, $\angle NGC=\angle NCG=90-B$. $\Rightarrow \angle NGE=90^{\circ}$. Similarly $\angle NFE=90^{\circ}$. Thus $NFEG$ is also cyclic.

The circumcircle of $EFG$ is same as circumcircle of quadrilateral $NFEG$ whose diameter is $NE$.

Update :

The above is how one would explain in detail. But to quickly identify that the statement holds true, recall that the nine-point circle is the unique circle that passes through the midpoints of three sides of a triangle, the feet of three altitudes and midpoints of the three segments joining each vertex to the orthocenter. $\odot(EFG)$ is indeed the nine-point circle (you also show its radius is $R/2$). Once we identify the center of $\odot(CFG)$, $N$, as midpoint of $CH$, we're done.

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    $\begingroup$ @Prags Yes, his answer is nice too. Since you marked this answer as accepted, I added an update. $\endgroup$
    – cosmo5
    Feb 10 at 10:25
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    $\begingroup$ @Prags good decision in the end :) $\endgroup$
    – Math Lover
    Feb 10 at 11:05
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Quadrilateral $ABGF$ is cyclic so

$\angle CFG = \angle ABC$ and $\angle CGF = \angle BAC$.

$GE$ is the median to the midpoint of hypotenuse in a right angled triangle so $GE = BE = EE, \angle BGE = \angle ABC$. Similarly $FE = AE = BE, \angle AFE = \angle BAC$.

So $\triangle EFG$ is an isosceles triangle with $FE = GE$. That leads to,

$\angle GFE = \angle EGF = 180^0 - (\angle ABC + \angle BAC) = \angle ACB$
$\angle FEG = 180^0 - 2 \angle ACB \ $.

$ \therefore \ \angle FNG = 2 \angle ACB$ and as $N$ is a point on one of the perpendicular bisectors, it has to be the circumcenter of $\triangle FCG$.

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  • $\begingroup$ Thanks! Would you like to chat? $\endgroup$
    – cosmo5
    Feb 10 at 11:29
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Probably the easiest:
Enough to show that $FNGE$ is cyclic $\iff$ $FNG=180-FEG$. Also $FEG=180-(FEA+GEB)$ and clearly because $FE$ and $GE$ are the medians of right triangles $AFB$,$AGB$ respectively, it's well-known that $FEA=180-2A$ and $GEB=180-2B$ so $FEG=180-(180-2A)-(180-2B)=180-2C$ and $FEG=2C$ is clear because $N$ is the circumcenter of $CFG$.

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