如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD. (1)求∠ABC的度数; (2)求证:如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD.(1)求∠ABC的度数;(2)求证:△CAF为等
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![如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD. (1)求∠ABC的度数; (2)求证:如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD.(1)求∠ABC的度数;(2)求证:△CAF为等](/uploads/image/z/1138652-44-2.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%88%A5BC%2CAB%3DAD%3DDC%2CAC%E2%8A%A5AB%2C%E5%BB%B6%E9%95%BFCB%E8%87%B3F%2C%E4%BD%BFBF%3DCD%EF%BC%8E+%EF%BC%881%EF%BC%89%E6%B1%82%E2%88%A0ABC%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B+%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%88%A5BC%2CAB%3DAD%3DDC%2CAC%E2%8A%A5AB%2C%E5%BB%B6%E9%95%BFCB%E8%87%B3F%2C%E4%BD%BFBF%3DCD%EF%BC%8E%EF%BC%881%EF%BC%89%E6%B1%82%E2%88%A0ABC%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3CAF%E4%B8%BA%E7%AD%89)
如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD. (1)求∠ABC的度数; (2)求证:如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD.(1)求∠ABC的度数;(2)求证:△CAF为等
如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD. (1)求∠ABC的度数; (2)求证:
如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD.
(1)求∠ABC的度数;
(2)求证:△CAF为等腰三角形.
如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD. (1)求∠ABC的度数; (2)求证:如图,在梯形ABCD中,AD∥BC,AB=AD=DC,AC⊥AB,延长CB至F,使BF=CD.(1)求∠ABC的度数;(2)求证:△CAF为等
设∠D=X°
因为ΔACD是等腰三角形
⑴所以∠CAD=﹙180°-X°﹚/2=90°-﹙X°/2﹚
所以∠BAD=∠CAD+∠BAC=90°-﹙X°/2﹚+90°=180°-﹙X°/2﹚
因为是等腰梯形,所以∠BAD=∠D=180°-﹙X°/2﹚=X°
解得X°=120°
所以∠ABC=180°-∠BAD=180°-120°=60°
⑵∠ABF=180°-∠ABC=180°-60°=120°
因为BF=CD=AB
所以ΔABF是等腰三角形
∠F=﹙180°-120°﹚/2=30°
∠ACB=180°-60°-90°=30°
所以∠F=ABC
三角形CAF的两个底角相等
所以△CAF是等腰△
1)∵AD∥BC,
∴∠DAC=∠ACB.
∵AD=DC,
∴∠DCA=∠DAC.
∴∠DCA=∠ACB=1 2 ∠DCB.
∵DC=AB,
∴∠DCB=∠ABC.
∴∠ACB=1 2 ∠ABC.
在△ACB中,∵AC⊥AB,
∴∠CAB=90°.
∴∠ACB+∠ABC=90°.
∴1 2 ∠ABC+∠ABC=...
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1)∵AD∥BC,
∴∠DAC=∠ACB.
∵AD=DC,
∴∠DCA=∠DAC.
∴∠DCA=∠ACB=1 2 ∠DCB.
∵DC=AB,
∴∠DCB=∠ABC.
∴∠ACB=1 2 ∠ABC.
在△ACB中,∵AC⊥AB,
∴∠CAB=90°.
∴∠ACB+∠ABC=90°.
∴1 2 ∠ABC+∠ABC=90°.
∴∠ABC=60°
2)证明:连接DB,
∵在梯形ABCD中,AB=DC,
∴AC=DB.
在四边形DBFA中,DA∥BF,DA=DC=BF,
∴四边形DBFA是平行四边形.
∴DB=AF,
∴AC=AF.
即△ACF为等腰三角形
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