(2009•重庆)已知:如图,在直角梯形ABCD中,AD‖BC,∠ABC=90°,DE⊥AC于点F,交(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求AE=AC
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![(2009•重庆)已知:如图,在直角梯形ABCD中,AD‖BC,∠ABC=90°,DE⊥AC于点F,交(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求AE=AC](/uploads/image/z/9864159-15-9.jpg?t=%282009%26%238226%3B%E9%87%8D%E5%BA%86%29%E5%B7%B2%E7%9F%A5%3A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%80%96BC%2C%E2%88%A0ABC%3D90%C2%B0%2CDE%E2%8A%A5AC%E4%BA%8E%E7%82%B9F%2C%E4%BA%A4%EF%BC%882009%26%238226%3B%E9%87%8D%E5%BA%86%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%88%A5BC%2C%E2%88%A0ABC%3D90%C2%B0%2CDE%E2%8A%A5AC%E4%BA%8E%E7%82%B9F%2C%E4%BA%A4BC%E4%BA%8E%E7%82%B9G%2C%E4%BA%A4AB%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9E%2C%E4%B8%94AE%3DAC%EF%BC%8E1.%E6%B1%82AE%3DAC)
(2009•重庆)已知:如图,在直角梯形ABCD中,AD‖BC,∠ABC=90°,DE⊥AC于点F,交(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求AE=AC
(2009•重庆)已知:如图,在直角梯形ABCD中,AD‖BC,∠ABC=90°,DE⊥AC于点F,交
(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.
1.求AE=AC
2.∠E=30,BC=3,求DC的长
(2009•重庆)已知:如图,在直角梯形ABCD中,AD‖BC,∠ABC=90°,DE⊥AC于点F,交(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求AE=AC
1.连接AG (BG=FG AG=AG ABG=AFG=90)
所以AB=AF BAG=CAG AG是BAC平分线
E=90-BGE=90-FGC=ACG BAG=CAG AG=AG
所以AEG全等ACG
AE=AC
2.BG=FG=x GC=2x BC=CG+BG=3x x=1
BAG=GAF=FAD=30
AG=AD=GC=2(这个好证) AGCD平行四边形 DE⊥AC
AGCD菱形 DC=2
AC在哪?
如图,连接EC,如AD=DC,且DF垂直于AC,可知FD是角ADC的平分线,故D;也是角AEC的平分线,且AE=EC;又已知AE=AC,所以AEC是等边三角形,可得角ACB=30°,故可得2DF=AD=2,得DF=1
怎么第一问在题设中有???