五边形ABCDE,∠ABC=∠AED=90° ,F为CD中点,CF=FD ,∠BAC=∠EAD, 求证FB=FE
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五边形ABCDE,∠ABC=∠AED=90° ,F为CD中点,CF=FD ,∠BAC=∠EAD, 求证FB=FE
![五边形ABCDE,∠ABC=∠AED=90° ,F为CD中点,CF=FD ,∠BAC=∠EAD, 求证FB=FE](/uploads/image/z/17860821-69-1.jpg?t=%E4%BA%94%E8%BE%B9%E5%BD%A2ABCDE%2C%E2%88%A0ABC%3D%E2%88%A0AED%3D90%C2%B0+%2CF%E4%B8%BACD%E4%B8%AD%E7%82%B9%2CCF%3DFD+%2C%E2%88%A0BAC%3D%E2%88%A0EAD%2C+%E6%B1%82%E8%AF%81FB%3DFE)
http://i159.photobucket.com/albums/t145/l421013/MATH2/FD-1.png
如图,已知五边形ABCDE中,∠ABC=∠AED=90°,∠BAC=∠EAD,F是CD中点,求证:FB=FE.
如图,在五边形ABCDE中,∠ABC=∠AED=90°,M是CD的中点,BM=EM,求证:∠BAC=∠EAD.
如图在五边形ABCDE中,角ABC=角AED=90度.BM=EM.M是CD中点.求证角BAC=角EAD
五边形ABCDE中,AB=AE,BC=DE,点F是CD的中点,且AF平分CD,求证:∠ABC=∠AED
如图,已知△ABC和△ADE中,∠ABC=∠AED=90°∠BAC=∠EAD,M为CD的中点,求证:MB=ME
五边形ABCDE中,已知AB=CD=AE=BC+DE=2,∠ABC=∠AED=90°,求五边形ABCDE的面积.
如图,五边形ABCDE中AB=AE,BC=ED,∠ABC=∠AED,点F是CD的中点
1、如图,在五边形ABCDE中,∠ABC=∠AED, ∠BCD=∠EDC,BC=DE,M为CD的中点,则AM垂直
(1)如图,在凸五边形ABCDE中,AB⊥BC,AE⊥ED,∠BAC=∠EAD,P是CD的中点,求证:PB=PE.(提示
如图,五边形ABCDE中,AB=AE,BC+DE=CD,∠ABC+∠AED=180°,连接AC、AD,求证:AD评分∠C
在五边形ABCDE中,AB=AE、BC+DE=CD,∠ABC+∠AED=180°. 求证:AD平分∠CDE.
(初二)在五边形ABCDE中,AB=AE、BC+DE=CD,∠ABC+∠AED=180°.求证:AD平分∠CDE.