Locate The Circumcenter

Assuming $AB$ and $AD$ are parallel to the $x$ - and $y$ -axis respectively, then we note that $EF$ is parallel to $AD$ and $BC$ . Since $\dfrac {HC}{HE} = \dfrac {HB}{HF} = \dfrac {BC}{EF} = \dfrac 21$ $\implies BC = AD = 2EF = 2\left(\dfrac {29}5 - \dfrac 95\right) = 8$ .

Since $G(1,5)$ is the midpoint of $AD$ , $\implies A(1,1)$ . Let $AB = a$ , then $B(1+a, 1)$ . Let $\angle DAC = \angle ACB = \angle ABH = \theta$ . Drop a perpendicular from $F$ to $AB$ at $I$ . Then we have:

$\begin{cases} \tan \theta = \dfrac {CD}{AD} = \dfrac a8 & ...(1) \\ \tan \theta = \dfrac {FI}{BI} = \dfrac {\frac 95-1}{1+a-\frac {17}5} = \dfrac 4{5a-12} & ...(2) \end{cases}$

$\begin{aligned} (1)=(2): \quad \quad \frac a8 & = \frac 4{5a-12} \\ 5a^2-12a - 32 & = 0 \\ (5a+8)(a-4) & = 0 \\ \implies a & = 4 & \small \color{#3D99F6} \text{Since }a > 0 \end{aligned}$

$\implies B(5,1)$ and since circumcenter is the meeting point of perpendicular side bisectors, $x_0 = \dfrac {x_A+x_B}2 = \dfrac {1+5}2 = 3$ . Using Pythagorean theorem , we have:

$\begin{aligned} \left(x_0-x_A\right)^2 + \left(y_0-y_A\right)^2 & = \left(x_E-x_0\right)^2 + \left(y_E-y_0\right)^2 \\ \left(3-1\right)^2 + \left(y_0-1\right)^2 & = \left(\frac {17}5 - 3 \right)^2 + \left(\frac {29}5-y_0\right)^2 \\ \implies y_0 & = 3 \end{aligned}$

$\implies x_0 + y_0 = 3+3=\boxed{6}$

(See above image for details)

There are a few aproaches to this but this is a suggested solution I have come up with.

First we will find the coordinates of point $A$

We have $HE=EC,HF=FB \Rightarrow \overrightarrow{EF}= \frac{1}{2}\overrightarrow{CB} = \frac{1}{2}\overrightarrow{DA} = \overrightarrow{GA}$

$\Rightarrow$ the coordinates of $A$ is $(1;1)$ .

We will find the coordinates of $B$

We have $EF \parallel BC, BC \perp AB \Rightarrow EF \perp AB$ . We also have $BH \perp AC \Rightarrow F$ is the orthocenter of $\Delta ABE$

We have: $\overrightarrow{AB} \perp \overrightarrow{EF}(0;-4)$ and passes point $A(1;1)$ so the equation of $AB$ is $y=1$

Similarly we have $\overrightarrow{BH} \perp \overrightarrow{AE}$ and passes point $F$ so the equation of $BH$ is $x+2y-7=0$

$BH \cap AB=B$ so the coordinates of $B$ is $(5;1)$ .

Now, let $O(x_0;y_0)$ be the circumcenter of $\Delta ABE$ , $EK$ be the diameter of $(O)$ .

We have $AF \perp EB, KB \perp EB \Rightarrow AF \parallel KB$ and $AK \perp AE, BF \perp AE \Rightarrow AK \parallel BF$ so we can imply that $AKBF$ is a parallelogram. Let $KF \cap AB=I \Rightarrow$ point $I$ is the midpoint of $AB$ . We have found the coordinates of points $A,B$ so we can find the coordinates of point $I$ that is $(3;1)$ .

We have $O$ is the midpoint of $EK$ and $I$ is the midpoint of $FK$ so $\overrightarrow{OI}= \frac{1}{2}\overrightarrow{EF}$ . We have had the coordinates of $E,F,I$ so we can imply the coordinates of $O$ that is $(3;3)$ .

In conclusion, $x_0+y_0=3+3= \boxed6$

We know EF is parallel to BC, and equals half of BC. From that we get AG = 4, which leads to A=(1,1). Then slope of AE is 2, and we know BC = 2 EF = 8, so AB = 8/2 = 4, which gives B = (5,1). Now we know the midpoint of AB to be (3,1), so the perpendicular bisector of AB is x = 3. To find the circumcenter of ABE, we need to find another perpendicular bisector, and find where it intersects with x = 3. For this we try side AE, by midpoint formula we get the midpoint of AE to be (11/5, 17/5), slope of its perpendicular bisector is -1/2, so the equation for AE's perpendicular bisector is y - 17/5 = -1/2 (x-11/5). Solve this together with x = 3, gives us y = 3. Thus the circumcenter is at (3,3), and answer is 3+3 = 6.

Notice that throughout the solution process, we didn't use the fact that BH is perpendicular to AC, it could be any point on diagonal AC.