1.The figure on the right is composed of square ABCD and tri

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1.The figure on the right is composed of square ABCD and triangle BCE,where ∠BEC is right angle．Suppose the length of CE is a,and the length of BE is b,then the distance between point A and line CE equals to ．
(be composed of 由…组成 right angle 直角 length 长度 distance 距离)
2.在△ABC中,AB＞BC,BD平分∠ABC,若BD将△ABC的周长分为4∶3的两部分,则△ABD与△BCD的面积比等于
Sorry啊,我不会放图,应该可以想象出来吧.
．
第一题，△BCE在正方形的外部。可以回答了吗

1.图不是不能想象,但是有歧义,你只要告诉我三角形BCE是在正方形的外部还是内部就可以了,你不会贴图,那补充提问会的吧?
2.答案是4：3么?
好了,可以回答了～
1.距离是a+b
首先作图,过A作CE的垂线,交CE于F,交CB于G,那么要求的就是AF的长度,然后过B作AF的垂线交AF于H
△CGF和△AGB都是直角三角形,而且有一组对顶角,所以∠BCE=∠FAB
又因为AB=AC,所以Rt△CBE≌Rt△ABH
所以AH=CE,BH=BE
BH是AF垂线,AF是CE垂线,∠CEB本身是直角,所以HBEF是矩形,也就是说：HF=BE
所以 AF=AH+HF=CE+BE=a+b
2.先画图,把AB画的比BC长就行.在AB上取点E,使得BE=BC,连接ED.因为BD是角平分线,所以△BDE≌△BDC.为了简便,现在在图上标出来,令AE=a,BE=BC=b,ED=DC=c,AD=d
S△CDB:S△ADB=CD:CA=c:d（因为这两个三角形的底都在AC上,又公用顶点B,所以等高,面积比就是两个底的长度比,接下来都是利用这个性质）
又S△CDB:S△ADB=S△BDE:S△ADB=BE:(AE+BE)=b:(a+b)
所以b:(a+b)=c:d,也就是：(a+b):b=d:c
又根据已知：(a+b+d):(b+c)=4:3
所以根据等比的性质：(a+b):b=d:c=4:3
而我们要求的正是S△ADB:S△CDB=d:c=4:3
其他回答的朋友若发现我上述有哪里犯错,欢迎指正～

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