请解答一下上面这道题---**Question:** 20. 如图, ⊙O 是 △ABC 的外接圆, 且 AB=AC, 作 AD⊥AB, 交 ⊙O 于点 D, 交 BC 延长线于点 E, 过点 B 作 ⊙O 的切线交 DA 的延长线于点 F. (1) 求证: BF=BE; (2) 若 ⊙O 的半径为 13, BC=24. 求 CE 的长. **Diagram Description:** * **Type:** Geometric figure. * **Elements:** * A circle with center O is shown. * A triangle ABC is inscribed in the circle (circumscribed circle of △ABC). * It is given that AB = AC. * A line segment AD is drawn such that AD is perpendicular to AB (∠DAB = 90 degrees). * The line AD intersects the circle at point D. * The line AD intersects the extension of BC at point E. * A line is drawn through B tangent to the circle at B. * The line DA is extended to intersect the tangent line at B at point F. * **Points:** A, B, C, D, E, F, O. * **Lines:** Sides of the triangle AB, BC, AC. Line segment AD. Line BE (extension of BC). Line AF (extension of AD). Tangent line BF at B. * **Relationships shown:** * Points A, B, C are on the circle. * Point O is the center of the circle. * AB = AC. * AD ⊥ AB. * D is on the circle. * E is on the extension of BC. * BF is tangent to the circle at B. * F is the intersection of the tangent line BF and the extension of DA.