对下题进行讲解的视频 如图，在▲ABC中，AB=AC, AC,AB 边上的高分别为BD，CE，且相交于点 O. 求证：▲ABC是等腰三角形. 解题思路为：看到AB=AC，由等边即可得等角，所以∠ABC=∠ACB。又由高可知互余，利用等角的余角相等得∠OCB=∠OBC，由等角对等边，即可得结论。 解题步骤为： 第一步：定等角 ∵AB=AC ∴∠ABC=∠ACB 第二步：等角转化 ∵BD,CE分别是AC,AB边上的高, ∴∠BDC=∠CEB=90° ∴∠DBC+∠DCB=90°，∠ECB+∠EBC=90° 第三步：等角转化 又∵∠EBC=∠DCB，∴∠DBC=∠ECB ∴∠ACB-∠ACE=∠ABC-∠ABD， 即 ∠OCB=∠OBC 第四步：得出结论 ∴OB=OC ∴▲OBC是等腰三角形.---Chart/Diagram Description: * **Type:** Geometric figure, specifically a triangle with intersecting line segments inside. * **Main Elements:** * **Points:** Points A, B, C, D, E, and O are labeled. * **Lines/Segments:** * A triangle ABC is formed by segments AB, BC, and AC. * Segment BD connects vertex B to a point D on side AC. * Segment CE connects vertex C to a point E on side AB. * Segments BD and CE intersect at point O, which is located inside the triangle. * **Labels:** Points are labeled A, B, C, D, E, and O. Point D is on segment AC. Point E is on segment AB. Point O is the intersection of segments BD and CE.