Question

如图,已知:△ABC和△ADE均为等边三角形,点D在BC上,DE与AC交与点F 求证：AE²=AF·AC

Answer 1

解：
∵∠ADE=∠C=60°
∠DAC=∠DAC
∴△ADF∽△ACD
∴AF/AD=AD/AC
∴AD²=AF·AC
∵AD=AE
∴AE²=AF·AC

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Answer 2

:△ABC和△ADE均为等边三角形
<BAD=<FAE <ABC=<AEF=60° <AFE=<ADB
△BAD∽△EAF AE/AB=AF/AD 又因为 AD=AE AB=AC
即 AE/AC=AF/AE
AE²=AF·AC