如圖,△ABC中,CE平分∠ACB交AB于E,過E作EF∥BC交∠ACD的平分線于F、EF交AC于M,若CM=5,則CE2+CF2=_.
如圖,△ABC中,CE平分∠ACB交AB于E,過E作EF∥BC交∠ACD的平分線于F、EF交AC于M,若CM=5,則CE2+CF2=______.
數(shù)學人氣:323 ℃時間:2019-08-18 00:21:16
優(yōu)質(zhì)解答
∵CE平分∠ACB交AB于E,CF平分∠ACD,∴∠1=∠2=12∠ACB,∠3=∠4=12∠ACD,∴∠2+∠3=12(∠ACB+∠ACD)=90°,∴△CEF是直角三角形,∵EF∥BC,∴∠1=∠5,∠4=∠F,∴∠2=∠5,∠3=∠F,∴EM=CM,CM=MF,∵CM=5,...
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