梯形ABCD AB平行DC AD=BC AE,BF分别是两腰上的高 且AE,BF相交与点O
设角BAE=X 角C=Y 找出并证明X与Y之间的关系式
人气:135 ℃ 时间:2019-08-20 11:16:41
解答
∵AB‖DC ∴∠EBA=∠C
又∵∠AEB=90 ∴∠C+∠BAE=90
故Y=90-X
(那个BF⊥AD的条件我怎么也用不上)
推荐
- 如图在梯形ABCD中AD平行于BC AB=CD AE平行于DC交BC于点E BF⊥AE于F 已知AD=2 BC=7 CF=4 且∠ABF=∠DCF求AB
- 梯形ABCD中,AD//BC,∠ABC=90°,F是AB延长线上一点,BF=AB,DF交AB于E,∠ADF=60°,AE=DC=2
- 如图,在梯形ABCD中,AD∥BC,AB=DC=AD,∠C=60°,AE⊥BD于E,AE=1.求梯形ABCD的高.
- 如图,在梯形ABCD中,AD∥BC,AB=DC=AD,∠C=60°,AE⊥BD于E,AE=1.求梯形ABCD的高.
- 在四边形ABCD中,AD‖BC,角ABC=角DCB,AB=DC,AE=DF.求证:BF=CE
- I did what everyone does who has no idea what to do with themselves when they got out of college and went on to graduate
- Tom has over ten books.(改同义句) Tom has ____ ____ ten books.
- 成语‘狡兔三窟’的来历,请告诉.
猜你喜欢