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There is given an acute triangle $ABC$ with $\angle ACB = 45^{\circ}$. Let $BCED$ and $ACFG$ be squares built on the sides of triangle $ABC$. Prove that the midpoint of the segment $DG$ is also the center of the escribed circle of the $ABC$ traingle.

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Man, this problem turned out to be much simpler that I expected. Notice that $$\angle \, GCA = 45^{\circ} = \angle \, BCD$$ and since by assumption $\angle \, ACB = 45^{\circ}$, angles $$\angle\, GCB = \angle \, GCA + \angle\, ACB = 45^{\circ} + 45^{\circ} = 90^{\circ}$$ $$\angle\, ACD = \angle \, ACB + \angle\, BCD = 45^{\circ} + 45^{\circ} = 90^{\circ}$$ which means that $GC$ is orthogonal to $BC$ and $DC$ is orthogonal to $AC$.

Let $O_B$ be the intersection point of the diagonals of square $ACFG$, so $O_B$ is the center of this square. Similarly, let $O_A$ be the intersection point of the diagonals of square $BCEF$, so $O_A$ is the center of this square. Consequently, point $O_B$ is the midpoint of segment $CG$ (diagonal of a square); point $O_A$ is the midpoint of segment $CD$ (diagonal of a square). Let $O$ be the midpoint of $DG$. Then in triangle $GDC$, segment $OO_A$ is parallel to $GC$ and half in length, as well as segment $OO_B$ is parallel to $DC$. Since $GC$ is orthogonal to $BC$ and $DC$ is orthogonal to $AC$, the segment $OO_B$ is orthogonal to $BC$ and $OO_A$ is orthogonal to $AC$. Since triangles $BO_AC$ and $AO_BC$ are right angled isosceles triangles and the lines $OO_A$ and $OO_B$ are respective altitudes from the right angled vertices to the hypotenuses, $OO_A$ and $OO_B$ are also orthogonal bisectors of edges $BC$ and $AC$ respectively. Hence their intersection point $O$ is the circumcenter of triangle $ABC$.

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I assume that the squares are built outside the triangle.

Like in the following picture:

The red point is the midpoint of DG

Now, if we think of all this stuff to be complex numbers, we get:

$$D = B + (-i) \cdot (C - B)$$ $$G = A + i \cdot (C-A)$$

Hence:

$$\frac{D+G}{2} = \frac{B +A}{2} +i\cdot \frac{(B-A)}{2}$$

So we have proved that the red point is on the perpendicular bisector To the segment $AB$. In addition, we have already proved that the altitude from the red point to $AB$ equals to $AB/2$

So if we denote the red point by $R$, We can conclude $\angle ARB = \pi/2$.

So we have: $\angle ARB = 2\angle ACB$ and in addition $R$ is on the perpendicular bisector of $AB$

So we can conclude that R is either the circumcenter of the triangle, or it's reflection with respect to $AB$.

Now it is enough to check the oriented angles:

In the above picture I assumed that the orientation of $\Delta ABC$ i counter clockwise, and in this case (mod $2\pi$): $$\angle ACB = \frac{\pi}4$$

$$\angle ARB = arg\left(\frac{B-R}{A-R}\right) = arg\left(\frac{B-\left(\frac{B +A}{2} +i\cdot \frac{(B-A)}{2}\right)}{A-\left(\frac{B +A}{2} +i\cdot \frac{(B-A)}{2}\right)}\right) = arg(\frac{1-i}{-1-i}) = arg(i) = \frac{\pi}2$$

So $R$ is on the perpendicular bisector to $AB$, and in addition $\angle ARB = 2\angle ACB$ (mod $2\pi$) and it is sufficient to conclude that $R$ is the circumcenter of the triangle, Since the circumcenter lies on the perpendicular bisector, and is the only point for which: $\angle ARB = 2\angle ACB$ (mod $2\pi$)

(There is another point on the perpendicular bisector for which: $\angle ARB = - 2\angle ACB$ (mod $2\pi$) but we have checked that the signes are compatible, so it is not that point.)

Hence, R is the circumcenter of $\Delta ABC$

Q.E.D.

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