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三角形ABC中，D、E分别是AB边AC边上的点，且AD=AE，BE=CD，求证AB=AC? - 知乎 因为 A D = A E AD=AE ，所以 A D E \triangle ADE 是等腰三角形，那么底角 ∠ A D E = ∠ ∠ A E D <90 ∘ \angle ADE=\angle\angle AED<90^\circ 。 于是它们的补角 ∠ B D E = ∠ C E D> 90 ∘ \angle BDE=\angle CED>90^\circ ，即此时的 B D E, C E D \triangle BDE,\triangle CED 都是钝角三角形。 又因为 B E = C D, D E = E D BE=CD,DE=ED ，所以 B D E ≅ C E D