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ABCD
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ABCD人物
ABCD介绍主角
不科学呀。这么好看的漫画,无论是画风还是男主人设都ABCD的👌k啊,这部漫画一开场,我就已经爱上了🙊
命啊_lx
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ABCD漫画作者
Egology(かひと)
ABCD配音
![](https://f2.kkmh.com/image/200702/V6RW60C5y.jpg)
奥特曼的小糖豆
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ABCD漫画人物
顺次连接四边形ABCD各边中点形成一个菱形,则原四边形对角线AC、BD的关系是 ...
AC=BD【解析】略
顺次连接四边形ABCD各来自边中点形成一个菱形,则原四边形对角线AC、BD的关系是 . ...
AC=BD根据菱形的性质来解答该题.菱形的四条边相等,故四边形的对角线就一定要相等.解:∵EFG...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
顺次连接四边形ABCD各边中点政达形成一个菱形,则原四边形对角线AC、BD的关系是 ...
AC=BD解析:略
顺次连接四边形ABCD各边中点形成一个菱形,则原四边形对角线AC、BD的关系是_...
AC=BD略
顺次连接四边形ABCD各边中点形成一个菱形,则原四边形对角线AC、BD的关系是 ...
AC=BD解析:略
如图,已知平行四边形ABCD中,AB=1/2AD,AB=AE=BF,探索EC与FD的位置关系
EC与FD的360问答位置关系是:互相垂直。异贵证明:设AD与EC的也山未军局较条载表交点为M,...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如图,矩形ABCD中,M是AD的中点.(1)求证:△ABM≌△DCM;(2)请你探索,当矩形ABCD中的一组邻边满足何种数量关系时,有BM⊥CM成立,观皮从缩守适她景尽政首说明你的理由.
试题**:(1)证明:∵矩形ABCD中,M360问答是AD的中点.∴AM=DM,∠A=∠D=90...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如图,在梯形ABCD中,AB∥CD,∠A+∠B=90º,E,F分别为AB,C来自D的中点。试探索EF,AB,CD之间的关系
2EF=AB-CD证明:作FM//360问答DA,交AB于M么,FN//CB交AB于N则∠FMN...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
在梯形ABCD中,AB平行CD,(AB大于CD)点E.F分别是***.CD的中点,若角A+角B=180度,试探索***.CD.EF的关系
过F做DA、CB的平行线交AB于M、N角MFN=90度而且DF=AM、CF=BN所以ME=EN所...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如图,在梯形ABCD中,AB平行CD,∠A+∠B=90°,E,F分别为AB,CD的中点,试探索EF,AB,CD之间的关系。
EF=(AB-CD)/2。过F作FM平行AD交AB于M,FN平行BC交AB于N。(下面简单地说一...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如来自图,已知平行四边形ABCD中,AB=1/2AD,AB=AE=BF,探索EC与FD的位置关系,并说明
EC与FD的位置关系是:互相垂直。证明:设AD与EC的交点为M,BC与FD的交点为补**西美扩销...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如图,已知平行四边形ABCD中,AB=1/2AD,AB=AE=BF,探索EC与FD的位置关系,并说明理由
EC与FD的位置关系是:互相是垂直。证明:设AD与EC的交点为M,BC与FD的交点为N,连结MN...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
如图,已知平来自行四边形ABCD中,AB=360问答二分之一AD,AB=AE=BE,探索EC与FD的位置关系,并说理由
EC与FD的位置关系示严步础该员笑是:互相垂直。证明:设AD与EC的交点为M,BC与FD的交点为...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
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