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D of E
已知线段AC与BD相交于点O,连接AB.DC.
E
为OB的中点,F为OC的中点,连接EF...
答:
(2)由②③能得到AB=CD,OB=OC,但不能证得△OAB≌△ODC,不能得到①中结论.解答:解:(1)①②→③或①③→②;①②→③ 证明如下:∵∠OEF=∠
OFE
∴OE=OF,∵E、F分别为OB、OC的中点 ∴OB=OC,在△OAB与△ODC中:∵∠A=∠
D
,∠AOB=∠DOC,OB=OC,∴△OAB≌△ODC(ASA),...
...
e
-mail ___ Helen ___ me. A.to;from B.from;to C.from;
of
D
...
答:
为你解答。(B)Please read this
e
-mail (from) Helen (to) me.解释:from表示寄信人, to表示收信人。(B)Lucy (wants) to pick the apples (on) the tree.解释:果实在树上,用on the tree。动物或人在树上,用in。(A)There are lots
of
things (to do) at school.解释:there ...
翻译《普鲁弗洛克情歌》
答:
(韵脚是a;bb,c,
d
,
e
;d;e--c.) I grow old ... I grow old ... 我老了... 我老了... I shall wear the bottoms
of
my trousers rolled. 我该把裤脚卷起了. Shall I part my hair behind? Do I dare to eat a peach? 我该把头发向后分吗? 我敢去吃桃子吗? I shall wear white flan...
...CD为弦,且CD垂直AB垂足为
E
,交圆O于点C和点
D
,
OF
垂直AC于点F,∠D...
答:
连接OC ∵AB为⊙O的直径∴∠ACB=90°∵∠
D
=30°∴∠A=30°,∠AOC=120°∴AB=2BC,AC=√3BC∵BC=1∴AB=2,AC=√3,AO=1∵
OF
⊥AC∴OF=1/2∴扇形AOC=π×1²×120°/360°=π/3 S△AOC=√3/4∴阴影面积=π/3-√3/4=(4π-3√3)/12 ...
...平行四边形ABCD的顶点A的坐标为(-2,0),点
D
的坐标为(0,2√3_百度...
答:
(2)①证明:∵A(-2,0),
D
(0,2 3),且E是AD的中点,∴E(-1, 3),AE=DE=2,OE=OA=2,∴△OAE是等边三角形,则∠AOE=∠AEO=60°;根据轴对称的性质知:∠AOE=∠EOF′,故∠EOF′=∠AEO=60°,即OF′∥AE,∴∠OF′E=∠DEH;∵∠OF′E=∠
OFE
=∠DGE,∴∠DGE=∠DEH...
a few simple forms
of
english poems翻译
答:
if we'
d
been better!Another simple form
of
poem that students can easily write is the cinquain, a poem made up of five lines. With these, students can convey a strong picture in just a few words. Look at the examples (
D
and
E
) on the top of the next page.D Brother...
instead和instead
of
的区别
视频时间 05:53
...点
D
、F分别在OA、OB上,OD=
OF
,点
E
、G在OC上,DE//FG,求证:DE=FG_百度...
答:
证明:连接DG,因为OC是角AOB的平分线,所以角DOG=角FOG,因为OD=
OF
OD=OD,所以三角形DOG和三角形FOG全等,所以DG=FG 角
D
GO=角FGO,所以角DGE=角FGE,因为DE平行FG,所以角FGE=角DEG,所以角DGE=角DEG,所以DG=DE,所以DE=FG
如图,在直角坐标系中,平行四边形AOC
D
的边OC在X轴上,边AD与与Y轴交与...
答:
∵OF=S,∴FD=10-S,AE=t,DE=12-t 又∵∠OEF=∠EDF.∴∠AEO+∠FED=180°-∠OEF,∠DEF+∠EFD=180°-∠EDF.∴∠AEO=∠EFD,∠A=∠EDF,∴△AEO∽△DFE,∴AEDF=AODE.∴t10-s=1012-t,100-10s=12t-t2,∴s=110t2-65t+10(0<t<12);(3分)(3)∠
OFE
>∠FDE=∠OEF...
如图,点A和点B分别在x轴y轴的正半轴上,AD平分∠BAO交y轴于
D
,OC⊥...
答:
0,8);(2)∵EF∥AB;OC⊥AB.∴OE⊥EF,则EF=√(
OF
²-OE²)=4.作EM垂直OF于M,由面积关系可知:EF*OE=OF*EM,4*3=5*EM, EM=12/5.则:CE=HE=OM=√(OE²-EM²)=9/5.所以,S⊿ACE/S⊿OAE=CE/EO=(9/5)/3=3/5.(同高三角形的面积比等底边之比)...
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