ABC is an acute triangle with \angle BCA = 35 ^\circ. Denote the circumcenter of ABC as O and the orthocenter of ABC as H. If AO=AH, what is the value of \angle ABC (in degrees)?

To solve this problem, we need to utilize some properties of triangles and their special points.

Let's start by understanding the relationship between the circumcenter and orthocenter in a triangle.

The circumcenter (O) is the center of the circumcircle, which is the circle passing through all three vertices of the triangle. It is located at the intersection of the perpendicular bisectors of the triangle's sides.

On the other hand, the orthocenter (H) is the point where the triangle's altitudes intersect. Altitudes are perpendicular lines drawn from each vertex to the opposite side.

Given that AO = AH, we can conclude that point A lies on both the circumcircle and altitude of triangle ABC.

The altitude from A intersects the circumcircle at point A and another point, let's call it X. Since AO = AX, it means that point X is the midpoint of side BC.

Now, let's examine the triangle ABC. We are given that \angle BCA = 35 degrees, and we know that \angle BAC is an acute angle because ABC is an acute triangle.

Since angle B is opposite side AC, and AX is perpendicular to AC, angle BAX is a right angle. Therefore, \angle BAX = 90 degrees.

Now, we can find the measure of \angle ABC by using the fact that the sum of the angles in a triangle is 180 degrees.

\angle ABC = 180 - (\angle BCA + \angle BAC)
\angle ABC = 180 - (35 + 90)
\angle ABC = 180 - 125
\angle ABC = 55 degrees

Therefore, the value of \angle ABC is 55 degrees.