Let $ABC$ be a triangle with $A=(2,3)$, $B=(6,-1)$ and $C=(4,7)$.
The middle points of the sides are the following:$M_{AB}=\left (\frac{2+6}{2} , \frac{3+(-1)}{2}\right )=\left (4 , 1\right )$, $M_{AC}=\left (\frac{2+4}{2} , \frac{3+7}{2}\right )=\left (3 \mid 5\right )$ and $M_{BC}=\left (\frac{6+4}{2} , \frac{(-1)+7}{2}\right )=\left (5 ,3\right )$.
We have that the equation of the circle $K$ that passes by the points $M_{AB}$, $M_{AC}$ and $M_{BC}$ is \begin{equation*}\left (x - \frac{17}{6}\right )^2 + \left (y - \frac{17}{6}\right )^2 = \frac{170}{36}\end{equation*}
To determine the intersection point of the three height I did the following:
$h_a$ is orthogonal to $\vec{BC}$.
The slope of $\vec{BC}$ is $\frac{7-(-1)}{4-6}=\frac{8}{-2}=-4$.
The slope of $h_a$ is therefore $\frac{-1}{-4}=\frac{1}{4}$.
The equation of $h_a$ is: \begin{equation*}h_a: y-3 = \frac{1}{4}(x-2) \Rightarrow y-3 = \frac{1}{4}x-\frac{1}{2} \Rightarrow y=\frac{1}{4}x+\frac{5}{2}\end{equation*}
$h_b$ is orthogonal to $\vec{CA}$.
The slope of $\vec{CA}$ is $\frac{3-7}{2-4}=\frac{-4}{-2}=2$.
the slope of $h_b$ is therefore $\frac{-1}{2}$.
The equation of $h_b$ is: \begin{equation*}h_a: y-(-1) = -\frac{1}{2}(x-6) \Rightarrow y+1 = -\frac{1}{2}x+3 \Rightarrow y = -\frac{1}{2}x+2\end{equation*}
For the intersection point we have $h_a=h_b$: \begin{equation*}\frac{1}{4}x+\frac{5}{2}=-\frac{1}{2}x+2 \Rightarrow \frac{3}{4}x=-\frac{1}{2} \Rightarrow x=-\frac{4}{3}\cdot \frac{1}{2} \Rightarrow x=-\frac{2}{3}\end{equation*}
Replacing this in $h_a$ we get: \begin{equation*}h_a: y=\frac{1}{4}\cdot \left (-\frac{2}{3}\right )+\frac{5}{2}= -\frac{1}{6}+\frac{5}{2}=\frac{14}{6}\end{equation*}
The intersection point is therefore \begin{equation*}H\left (-\frac{2}{3};\frac{14}{6}\right )\end{equation*}
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How could we show that $K$ goes through the middle point of $\overline{CH}$ and through the foot of the perpendicular from C to segment AB?
Could you give me a hint?
HINT: the intersection Point of the hights is H $$(-7/3;7/3)$$ and then we have $$\left(-2/3-7/6\right)^2+\left(7/3-7/6\right)^2=\left(\frac{85}{18}\right)^2$$