# Show circle through feet of two altitudes and midpoint of third side passes through center of another circle

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 $$\triangle CFG$$. 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. $$\angle HFC=90^{\circ}=\angle HGC$$. 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.

• @Prags Yes, his answer is nice too. Since you marked this answer as accepted, I added an update. Feb 10 at 10:25
• @Prags good decision in the end :) Feb 10 at 11:05

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 $$\triangle FCG$$.

• Thanks! Would you like to chat? Feb 10 at 11:29

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 $$CFG$$.