如图,AD//BC,AE,BE分别为DAB,ABC的平分线,点E在CD上,求证:AB=BC+AD.
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![如图,AD//BC,AE,BE分别为DAB,ABC的平分线,点E在CD上,求证:AB=BC+AD.](/uploads/image/z/6632480-56-0.jpg?t=%E5%A6%82%E5%9B%BE%2CAD%2F%2FBC%2CAE%2CBE%E5%88%86%E5%88%AB%E4%B8%BA%26%2361648%3BDAB%2C%26%2361648%3BABC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E7%82%B9E%E5%9C%A8CD%E4%B8%8A%2C%E6%B1%82%E8%AF%81%3AAB%3DBC%2BAD.)
如图,AD//BC,AE,BE分别为DAB,ABC的平分线,点E在CD上,求证:AB=BC+AD.
如图,AD//BC,AE,BE分别为DAB,ABC的平分线,点E在CD上,求证:AB=BC+AD.
如图,AD//BC,AE,BE分别为DAB,ABC的平分线,点E在CD上,求证:AB=BC+AD.
过E做AD的平行线交AB于F,则EF也平行于BC,∠DAE=∠AEF=∠EAF,所以三角形AEF为等腰三角形,所以AF=EF.同理,BF=EF,所以AF=EF=BF,又因为AD//BC//EF,所以DE=CE.在梯形ABCD中,所以有AD+BC=2EF,又AB=AF+EF=2EF.所以AB=BC+AD.得证!