Question

在△ABC中,∠C=90°,CA=CB,M为CB中点,ME⊥AB于E,连接CE,求sin∠ACE

Answer 1

做高CD,连接DM交CE 于点O,
由已知,∠C=90°,CA=CB,M为CB中点,ME⊥AB于E可得：
DM//AC,DM = CM ,DE = ME ,ME//CD,CD=2ME=DE
所以,∠ACE = ∠DOE =∠COM ,EO/OC=1/2
CE?= CD?+ DE?= 5DE?,
CE =√5DE ,CO = 2CE/3= 2√5DE /3
DM=√2DE = CM ,(OM = DM/3 = √2DE /3)
所以,sin∠ACE = sin∠COM = CM/CO
= √2DE /(2√5DE /3) = (3√10)/10