如图,在四边形ABCD中,∠ABC=90°,CD⊥AD,AD²+CD²=2AB².(1)求证:AB=BC;(2)当BE⊥AD于点E时,求证:BE=AE+CD.
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![如图,在四边形ABCD中,∠ABC=90°,CD⊥AD,AD²+CD²=2AB².(1)求证:AB=BC;(2)当BE⊥AD于点E时,求证:BE=AE+CD.](/uploads/image/z/7165351-55-1.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E4%B8%AD%2C%E2%88%A0ABC%3D90%C2%B0%2CCD%E2%8A%A5AD%2CAD%26%23178%3B%2BCD%26%23178%3B%3D2AB%26%23178%3B.%281%29%E6%B1%82%E8%AF%81%EF%BC%9AAB%3DBC%3B%282%29%E5%BD%93BE%E2%8A%A5AD%E4%BA%8E%E7%82%B9E%E6%97%B6%2C%E6%B1%82%E8%AF%81%EF%BC%9ABE%3DAE%2BCD.)
如图,在四边形ABCD中,∠ABC=90°,CD⊥AD,AD²+CD²=2AB².(1)求证:AB=BC;(2)当BE⊥AD于点E时,求证:BE=AE+CD.
如图,在四边形ABCD中,∠ABC=90°,CD⊥AD,AD²+CD²=2AB².
(1)求证:AB=BC;
(2)当BE⊥AD于点E时,求证:BE=AE+CD.
如图,在四边形ABCD中,∠ABC=90°,CD⊥AD,AD²+CD²=2AB².(1)求证:AB=BC;(2)当BE⊥AD于点E时,求证:BE=AE+CD.
(1) ∵ AB² +BC² =AC² =AD² +CD² = 2AB²
∴ AB² +BC² =AB² +AB²
∴ AB=BC
(2) 如图所示从点B作垂直于DC直线交于F则有
BE=DF
∠1+∠2+∠3=90º=∠4+∠5 -------①
∵∠5+∠6+∠7 =180º ∠6=∠2=45º
∠7=90º-∠3
∴ ∠5+∠6+90º -∠3=180º
∴∠5=90º+∠3-∠6
=45º+∠3 代入式①
∠1+45+∠3=90º=∠4+∠45º+∠3
∴ ∠1=∠4
又 ∵AB=BC
∴ΔABE≌ΔCBF
∴AE=CF=DF-CD=BE-CD
∴BE=AE+CD
1)AD^2+CD^2=AC^2=AB^2+BC^2=2AB^2,所以AB^2=BC^2,所以AB=BC
2)从C作BE垂线,垂足为F,则CDEF为矩形,三角形CBF、BEA全等。因此可知BE=CD+AE