Re: [幾何] 一題幾何證明請教~
: 想請教大師指點
: 謝謝您，謝謝
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在 CF 上取 B' 使 BF = FB', 連 AB'
易證 △ABF 全等於 △AB'F, 故 ∠AB'F = 2∠C
再由外角關係推得 ∠B'AC = ∠C, 即 B'C = AB' = AB
所以 BC = BF + FB' + B'C = 2BF + AB = 2BF + 2AD
BE = BC/2 = BF + AD
AD = BE - BF = EF #
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思路是從 2∠B = ∠C 開始的, 為了把這兩個條件湊起來才把 △ABF 往右「翻」
然後為了思考怎麼湊出 AD = EF 把一些邊長關係拿來湊就會容易湊出 BC 那一條
D 跟 E 在這裡都只是當做一半的條件來用而已, 關鍵的還是 △ABF 那一翻
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