三角形双角平分线---Here is the extracted content from the image: **Title:** 三角形双角平分线模型 模型精讲 **Table:** | 类型 | 图示 | 条件 | 结论 | 巧记 | | :--------------- | :------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | :---------------------- | :------------------------ | :-------------------- | | 双内角平分线 | **Diagram Description:** A triangle ABC with point D inside. BD bisects angle ABC (marked with double arcs). CD bisects angle ACB (marked with single dots). | BD 平分∠ABC, CD 平分∠ACB | ∠D = 90° + (1/2)∠A | 内内等于 90°加一半 | | 一内一外角平分线 | **Diagram Description:** Triangle ABC with side BC extended to E. BD bisects angle ABC (marked with double arcs). CD bisects exterior angle ACE (marked with single dots). Point D is outside the triangle. | BD 平分∠ABC, CD 平分∠ACE | ∠D = (1/2)∠A | 一内一外等于一半 | | 双外角平分线 | **Diagram Description:** Triangle ABC with sides AB and AC extended to E and F respectively. BD bisects exterior angle EBC (marked with double arcs). CD bisects exterior angle FCB (marked with single dots). Point D is outside the triangle. | BD 平分∠EBC, CD 平分∠FCB | ∠D = 90° - (1/2)∠A | 外外等于 90°减一半 | **模型证明 (运用定理, 证明模型结论)** **双内角平分线** ∵ BD 平分 ∠ABC, CD 平分 ∠ACB, ∴ ∠ABD = ∠CBD = (1/2)∠ABC, ∠ACD = ∠BCD = (1/2)∠ACB. ∵ ∠D + ∠CBD + ∠BCD = 180° ①, ∠A + ∠ABC + ∠ACB = 180° ②, 由 ① × 2 - ② 得 2∠D - ∠A = 180°, ∴ ∠D = 90° + (1/2)∠A. **一内一外角平分线** ∵ BD 平分 ∠ABC, CD 平分 ∠ACE, ∴ ∠ABD = ∠CBD = (1/2)∠ABC, ∠ACD = ∠ECD = (1/2)∠ACE. ∵ ∠ACE = ∠A + ∠ABC ①, ∠ECD = ∠D + ∠CBD ②, 由 ① - ② × 2 得 ∠A - 2∠D = 0, ∴ ∠D = (1/2)∠A. **双外角平分线** ∵ BD 平分 ∠EBC, CD 平分 ∠FCB, ∴ ∠EBD = ∠CBD = (1/2)∠EBC, ∠FCD = ∠BCD = (1/2)∠FCB. ∵ ∠D + ∠CBD + ∠BCD = 180°, ∴ ∠D + (1/2)(∠EBC + ∠FCB) = 180°. 由 A 字模型可知 ∠EBC + ∠FCB = 180° + ∠A, ∴ ∠D + (1/2)(180° + ∠A) = 180°, ∴ ∠D = 90° - (1/2)∠A.