如图,点E、F分别在正方形ABCD的边DA,AB上,且BE⊥CF于点G.
![](https://img.xkw.com/dksih/QBM/2022/3/9/2932746499391488/3058353038753792/STEM/5d036c74c1ba4eb99f5fde0929f37352.png?resizew=125)
(1)求证:△ABE≌△BCF;
(2)若四边形AECF的面积为12,求BC的长.
![](https://img.xkw.com/dksih/QBM/2022/3/9/2932746499391488/3058353038753792/STEM/5d036c74c1ba4eb99f5fde0929f37352.png?resizew=125)
(1)求证:△ABE≌△BCF;
(2)若四边形AECF的面积为12,求BC的长.
更新时间:2022-09-03 09:38:05
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【推荐1】如图,平面直角坐标系中,点A(0,a)在y轴正半轴上,点B(0,b)(b≤0),点C(c,0)在x轴正半轴上,且a2﹣2ab+b2﹣c2=0.
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(3)如图2,当b<0时,点D在x轴正半轴上点C的右侧,且CD=OB,连接AD,射线BC交AD于点E.当点B在y轴负半轴上运动时,∠AEB的度数是否为定值?如果是,请求出
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【推荐2】在
中,
,
,直线
经过点A,过点B、C分别作
的垂线,垂足分别为点D、E.
![](https://img.xkw.com/dksih/QBM/editorImg/2023/3/3/f4959855-b034-4500-b400-edacd448cc78.png?resizew=337)
(1)规律探究:
(I)如图1,请探究线段
、
和
的数量关系并说明理由;
(II)如图2,当直线
绕点
旋转到图2的位置时,请再探究线段
、
和
的数量关系并说明理由;
(2)尝试应用:在图2中,延长线段
交线段
于点
,若
,
,
,求
.
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(1)规律探究:
(I)如图1,请探究线段
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d40b319212a7e7528b053e1c7097e966.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4eedae8d316c76e3d0b451256de03fb9.png)
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(II)如图2,当直线
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0f85fca60a11e1af2bf50138d0e3fe62.png)
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(2)尝试应用:在图2中,延长线段
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d40b319212a7e7528b053e1c7097e966.png)
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【推荐1】如图,正方形ABCD中,对角线AC与BD相交于O点,作
分别交AB、BC于点E、F,若AE=4,CF=3,则EF的长是多少?
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【推荐2】如图,O为平行四边形ABCD的对称中心,对角线AC⊥AB,过点O作直线
,分别交AD,BC于E,F,连接AF,CE.
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(1)证明:四边形AFCE是菱形;
(2)若四边形AFCE是正方形且BC=6,求AB的长.
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解答题-证明题
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【推荐1】请阅读如下材料.
如图,已知正方形ABCD的对角线AC、BD于点O,E是AC上一点,AG⊥BE,垂足为G.求证:OE=OF.
证明:∵四边形ABCD是正方形.
∴∠BOE=∠AOF=90°,且OA=OE.
又∵AG⊥BE,∴∠1+∠3=90°=∠2+∠3,即∠1=∠2.
∴Rt△BOE≌Rt△AOF,∴OE=OF.
![](https://img.xkw.com/dksih/QBM/2017/11/10/1814322111635456/1816174663049216/STEM/9bbe87a6f75741a58b33a639fd2db300.png?resizew=116)
⑴根据你的理解,上述证明思路的核心是利用 使问题得以解决,而证明过程中的关键是证出 .
⑵若上述命题改为:点E在AC的延长线上,AG⊥BE交EB的延长线于点G,延长AG交DB的延长线于点F,如图,其他条件不变.
求证:OF=OE.
如图,已知正方形ABCD的对角线AC、BD于点O,E是AC上一点,AG⊥BE,垂足为G.求证:OE=OF.
证明:∵四边形ABCD是正方形.
∴∠BOE=∠AOF=90°,且OA=OE.
又∵AG⊥BE,∴∠1+∠3=90°=∠2+∠3,即∠1=∠2.
∴Rt△BOE≌Rt△AOF,∴OE=OF.
![](https://img.xkw.com/dksih/QBM/2017/11/10/1814322111635456/1816174663049216/STEM/9bbe87a6f75741a58b33a639fd2db300.png?resizew=116)
⑴根据你的理解,上述证明思路的核心是利用 使问题得以解决,而证明过程中的关键是证出 .
⑵若上述命题改为:点E在AC的延长线上,AG⊥BE交EB的延长线于点G,延长AG交DB的延长线于点F,如图,其他条件不变.
求证:OF=OE.
![](https://img.xkw.com/dksih/QBM/2017/11/10/1814322111635456/1816174663049216/STEM/7198c167103b43a3ae0d300cfb2e39b2.png?resizew=141)
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【推荐2】如图,正方形
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