下面这个结论是 Andrew Jobbings 在 2011 年指出的：
AB 是圆 O 的一条直径， CD 、 EF 是两条垂直于 AB 的弦，并且以 CD 为直径的半圆和以 EF 为直径的半圆正好切于点 T 。那么，两个半圆的面积之和一定等于圆 O 的面积的一半。
你能证明这个结论吗？
下面是 cut-the-knot 网站上给出的一种证明方法。如果把两个半圆的半径分别记作 r1 和 r2 ，把整个大圆的半径记作 R ，那么我们只需要证明
π r12 / 2 + π r22 / 2 = π R2 / 2
即
r12 + r22 = R2
由于 △MCO 是直角三角形，由勾股定理可知
MO2 + MC2 = OC2
其中 MC 就是其中左边那个半圆的半径， OC 就是整个圆 O 的半径，因而我们只需要证明 MO 就是右边那个半圆的半径即可。现在，连接 CT 、 FT ，你会发现 △MCT 和 △NFT 都是等腰直角三角形，并且 C 、 T 、 F 在一条直线上。由于圆心角的度数是圆周角的两倍，因此 ∠COE = 2 ∠CFE = 90° ，由此可知 ∠1 + ∠2 = 90° 。而 ∠2 + ∠3 = 90° ，于是 ∠1 = ∠3 。根据同样的道理，我们还可以得到 ∠2 = ∠4 。再加上整个大圆的半径 OC = OE ，就足以判定 △MCO 和 △NOE 全等了。这说明 MO = NE ，而后者正是右边那个半圆的半径。
第一次拿到沙发
相当于求等腰梯形的外接圆直径
主要是证明∠COE=90°，因此CE²=2R²。∠CTE也是直角，所以CE² =CT²+ET²=2(r₁²+r₂²)。这题难度一般？
学习来了. 好久没来逛了.
这个真是太简单了
之前的高大上数学理论怎么降格为初中数学了呢？
that’s so easy. it took me less than 30 seconds to prove it.
CMO OBE 全等，嗯 考试就用这个
初一初二程度的水题..
当然我是说玩竞赛的..
这是我第一个看完的题目，之前的我根本都看不完！！！太难
连接[C,E]; [T,C]; [T,E]; 然后盯着图看就证完了…
就是两45RT里边，两斜和[C,E]组成新的RT； 由此[C,E]=sqrt(2)*sqrt(a^2+b^2)；
又在三角形(OCE)里面，目测得[C,E]=sqrt(2)*r；
由此r^2=a^2+b^2;
证明(COE)是RT，看CE对的两圆心角即可
目测得[C,E]=sqrt(2)*r；
谁告诉你角COE是直角的？虽然确实是直角，但是需要证明啊
我表示，刚看到回复补充的那句话。。。。
关键是CTF在同一直线上，这个是怎么说明的？
对顶角相等 都为 45度
任意一个圆 某段圆弧的圆心角度数 等于 相应圆周角度数的2倍
呃为什么现在都是放些初中的家庭作业题…
真的是太妙了，唉，我还要继续进步啊，希望以后多出这些简单的，让更多的小学生认识你
是不是可以这么想：当两个半圆中的任意一个无限趋近于0，另一个半圆的直径将会无限的趋近于大圆直径，当其中一个半圆的面积为0时，另一个半圆的直径就是大圆的直径，所以两个小圆的半径之和等于大圆的一半。文科数学没学过这种极限的东西，不知道能不能这样证明。求大神指点。
圆心角和圆周角的概念早都还给老师了，更别说知道圆心角度数是圆周角的二倍了。不过还是证明出来了。
设圆M半径是R1,圆N的半径是R2,且OT为x，
然后三角形CMO和三角形ONE都为直角三角形，且CO=OE，那么R1的平方+(R1-x)的平方=CO的平方=OE的平方=R2的平方+(R2+x)的平方。根据此式可推导出x=R1-R2，所以其实OM=R2，在三角形CMO中可以得到R1平方+R2平方=R平方
wow牛B
虽然辅助线一出来就能看出来，但不得不感叹相当的精巧啊~
我想出来证∠COE=∠CNE=90°的方法,没有楼主的简洁,
最后证△MCO 和 △NOE全等基本一样, 当然证全等还可以只用一组锐角相等因为斜边相等都是圆半径
连接
OC, OE, OD, OF, CF, DE
CNF, DNE 两组三点共线且CF=DE, CF⊥DE
△COF≌△DOE
∠OEN=∠OCN
CEON四点共圆
∠COE=∠CNE=90°
在没想出几何方法前想到一种暴力的三角方法法证明∠COE = π/2
首先M,N必须分居圆心O的一两侧, 这直观上是显然的, 证明也不难:
比如假设M在O的右侧并且向右侧画半径为r1的半圆, 那么
OM = √(r^2 – r1^2) = √[(r + r1)(r – r1)] > √(r – r1)^2 = r – r1, 即
OM + r1 > r T会到AB延长线上与题意不符
所以可设∠EOB = α, ∠COB = β, 并且一定有 0 < α < π/2, π/2 < β < π
r1 = r sin α, r2 = r sin β
由两圆相切的性质MN = r1 + r2, 但另一方面 MN = r cos α – r cos β (注意是减号因为cos β < 0)
所以有 sin α + sin β = cos α – cos β
等式两边分别用和差化积公式:
2sin[(α + β) / 2]cos[(α – β) / 2] = 2sin[(α + β) / 2]sin[(β – α ) / 2]
由于有 π/2 < α + β < 3π/2
π/4 < (α + β) / 2 < 3π/4
sin[(α + β) / 2] ≠ 0
cos[(α – β) / 2] = sin[(β – α ) / 2]
sin[(β – α ) / 2] – cos[(β – α) / 2] = 0
√2 sin[(β – α ) / 2 – π/4] = 0
由于 0 < (β – α) / 2 < π / 2
(β – α) / 2 = π/4
∠COE = β – α = π/2 得证
这里后面当然可以用全等但是也可以用三角暴力到底:
由于有, β – α = π/2
sin β = sin (α + π/2) = cos α
两半圆面积的和 = π/2 (r1^2 + r2^2) = πr^2/2(sin^2 α + sin^2 β) = πr^2/2(sin^2 α + cos^2 α)
= πr^2/2
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