如图,△ABC是圆O的内接三角形,∠BAC的平分线交BC于D,交圆O于E。求证:AB·AC=AD^2+BD·DC

2025-05-19 10:05:43
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回答(1):

∵∠ABD=∠ABC=∠AEC,∠BAD=∠EAC
∴△ABD∽△AEC
∴AB/AE=AD/AC
∴AB·AC=AD·AE=AD(AD+DE)=AD^2+AD·DE
∵∠ADB=∠CDE,∠ABD=∠ABC=∠AEC=∠CDE
∴△ABD∽△CDE
∴AD/CD=BD/DE
∴AD·DE=BD·CD
∴AB·AC=AD^2+BD·CD