1. **Problem statement:** In triangle ABC, point O lies inside the triangle such that line segment OA bisects angle A. Given that \(\angle ABO = \angle OCA\), prove that \(OB = OC\). 2. **Step 1: Understand the given information.** - OA bisects \(\angle A\), so \(\angle BAO = \angle CAO\). - \(\angle ABO = \angle OCA\). 3. **Step 2: Use the angle bisector property.** Since OA bisects \(\angle A\), by the Angle Bisector Theorem, we have: $$\frac{OB}{OC} = \frac{AB}{AC}$$ if O lies on BC, but here O is inside the triangle, so we need a different approach. 4. **Step 3: Consider triangles ABO and ACO.** - In \(\triangle ABO\) and \(\triangle ACO\): - \(\angle BAO = \angle CAO\) (since OA bisects \(\angle A\)) - \(\angle ABO = \angle OCA\) (given) - Side AO is common. 5. **Step 4: Use the Angle-Side-Angle (ASA) criterion.** In \(\triangle ABO\) and \(\triangle ACO\), two angles and the included side AO are equal, so: $$\triangle ABO \cong \triangle ACO$$ 6. **Step 5: Conclude the equality of sides.** By congruence, corresponding sides are equal, so: $$OB = OC$$ **Final answer:** \(OB = OC\) is proven by ASA congruence of triangles ABO and ACO.