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

45

correct?

To find the value of ∠ABC in degrees, we need to use the given information that AO = AH.

Here is a step-by-step explanation of how to get the answer:

1. Recall that the circumcenter of a triangle is the point where the perpendicular bisectors of the sides of the triangle meet. Let's call the midpoint of side BC as M.

2. Since AO = AH, it means that O lies on the perpendicular bisector of BC. Therefore, O is equidistant from points B and C. This implies that O lies on the line perpendicular to BC passing through its midpoint M.

3. Since O is on the perpendicular bisector of BC, this means that OA = OM. Therefore, triangle OAM is an isosceles triangle.

4. In an isosceles triangle, the angles opposite the equal sides are equal. So, we can conclude that ∠MOA = ∠MAO.

5. Since ∠BCA is given as 35°, we know that ∠MAO + ∠BCA = 90° (because it's a right angle). Therefore, ∠MAO = 90° - ∠BCA = 90° - 35° = 55°.

6. In triangle AOB, ∠OAB and ∠OBA are equal because AO = BO (since O is the circumcenter and OB is a radius of the circumcircle).

7. So, ∠OBA = ∠OAB = 55°.

8. In triangle ABC, ∠ABC + ∠ACB + ∠BCA = 180° (Sum of angles in a triangle).

9. Therefore, ∠ABC + 90° + 35° = 180° (Since ∠ACB is 90° as H is the orthocenter).

10. Simplifying the equation, we get ∠ABC = 180° - 90° - 35° = 55°.

So, the value of ∠ABC is 55 degrees.