Find the $ACB$ Angle of Triangle ABC

[maxkor]

The triangle ABC is given. Find the $ACB$ angle if the orthocenter of this triangle belongs to the circumscribed circle of the triangle $AOB$, where O is the center of the circumscribed circle of the $ABC$ triangle.

 

[jvicens]

Hello,

To find the $ACB$ angle in this scenario, we can use the fact that the orthocenter of a triangle is the intersection of its altitudes. In this case, since the orthocenter belongs to the circumscribed circle of triangle $AOB$, it must also lie on the perpendicular bisectors of sides $AO$ and $BO$. This means that the orthocenter is equidistant from points $A$ and $O$, as well as points $B$ and $O$.

Using this information, we can construct a right triangle with sides $OA$, $OB$, and $AB$. The angle $C$ of triangle $ABC$ will be the same as the angle $O$ of this constructed right triangle. This is because the angle $O$ is subtended by the same arc $ACB$ as the angle $C$.

Since we know that the orthocenter is equidistant from points $A$ and $O$, we can use the Pythagorean theorem to find the length of the altitude from $A$ to side $BC$. Let's call this altitude $h$.

$OA^2 = h^2 + (\frac{AB}{2})^2$

$OB^2 = h^2 + (\frac{AB}{2})^2$

Subtracting these two equations, we get:

$OA^2 - OB^2 = 0$

This tells us that $OA = OB$, which means that the constructed right triangle is actually an isosceles triangle. Therefore, the angle $O$ must be equal to the angle $C$. Thus, the $ACB$ angle is equal to the angle $O$ and we can find its measure by dividing the arc $ACB$ by 2.

I hope this helps. Let me know if you have any further questions.