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Sep 15th 2013, 15:09

延長 AE 相交 BC 於 G, 則三角形 ABE 及 GBE 全等 (ASA), 所以 E 為 AG 的中點.
BG = BA = 12, 所以 CG = BC - BG = 14 - 12 = 2.
同理, 延長 AD 相交 BC 於 F, 則三角形 ACD 及 FCD 全等, 所以 D 為 AF 的中點,
CA = CF = 10, 所以 FG = CF - CG = 10 - 2 = 8.
於三角形 AFG, 因為 D, E 是 AF, AG 的中點, 且 FG = 8,
依中點定理, DE = FG/2 = 8/2 = 4.

答：DE = 4

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