如图,AC⊥BC,AC=BC,D为AB上一点,BE⊥CD于E,AF⊥DC交CD延长线于点F,BE=28,AF=12,求EF的长.
如图,AC⊥BC,AC=BC,D为AB上一点,BE⊥CD于E,AF⊥DC交CD延长线于点F,BE=28,AF=12,求EF的长.
数学人气:679 ℃时间:2019-10-10 06:08:51
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∵AC⊥BC,BE⊥CD,AF⊥DC,∴∠F=∠BEC=∠ACB=90°,∴∠ACF+∠BCF=90°,∠BCF+∠CBE=90°,∴∠ACF=∠CBE.在△ACF和△CBE中∠F=∠BEC∠ACF=∠CBEAC=CB,∴△ACF≌△CBE(SAS),∴CF=BE,AF=CE.∵BE=28,AF=...
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