It is not difficult to understand the method of yll
I will post HKCEE format later
[問題]Prove arc ...
由 --- 於 星期五 五月 16, 2003 10:51 pm
siuhochung 寫到:Meowth 寫到:(1) I'm not interested in geometry.
(2) Plane geometry won't be helpful for engineering mathematics or discrete math.
志於工數或純數者,鑽研Plane geometry 這門2千年古學幫助不大.
(1) I'm not interested in discrete math, and know nothing about engineering mathematics.
(2) Temporarily, discrete math and engineering mathematics will never help me to solve plane geometry questions.
中學階段的學生，鑽研plane geometry 這門2千年古學有莫大益處
對志於IMO金牌的Kang, geometry當然重要.
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--- - 訪客
由 ---- 於 星期五 五月 16, 2003 10:56 pm
Meowth 寫到:siuhochung 寫到:Meowth 寫到:(1) I'm not interested in geometry.
(2) Plane geometry won't be helpful for engineering mathematics or discrete math.
志於工數或純數者,鑽研Plane geometry 這門2千年古學幫助不大.
(1) I'm not interested in discrete math, and know nothing about engineering mathematics.
(2) Temporarily, discrete math and engineering mathematics will never help me to solve plane geometry questions.
中學階段的學生，鑽研plane geometry 這門2千年古學有莫大益處
對志於IMO金牌的Kang, geometry當然重要.
我能夠當上香港代表隊才算吧，就算當上代表隊，得到銅牌已經非常滿足了...
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---- - 訪客
由 Raceleader 於 星期五 五月 16, 2003 11:28 pm
連BC。
∠DXC=180°-∠DXA=180°-90°=90° (直線上的鄰角)
∠XCB+∠XBC=∠DXC=90° (三角形外角)
弧AB:弧CD=∠ACB:∠CBD (弧長與圓周角成比例)
∴弧AB:弧CD:半圓周=∠ACB:∠CBD:90° (半圓上的圓周角)
∴弧AB=半圓周(∠ACB/90°)
∴弧CD=半圓周(∠CBD/90°)
∴弧AB+弧CD=半圓周(∠ACB/90°)+半圓周(∠CBD/90°)=半圓周(∠ACB+∠CBD)/90°=半圓周(90°)/90°=半圓周
∴弧AD+弧BC=圓周-(弧AB+弧CD)=圓周-半圓周=半圓周
∴弧AB+弧CD=弧AD+弧BC
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Raceleader - 訪客
由 Searchtruth 於 星期五 五月 16, 2003 11:50 pm
我個人認為...數學真ㄉ想學好...
學校ㄉ是決對不夠ㄉ!!
因為...那些是符合中等程度ㄉ人ㄉ內容...
想要在哪科強...就自行處理@@"
說真ㄉ...我自認我"幾何"最爛= ="
不過我還是蠻喜歡作幾何題...
只要是數學~我什麼都好= =+
學校ㄉ是決對不夠ㄉ!!
因為...那些是符合中等程度ㄉ人ㄉ內容...
想要在哪科強...就自行處理@@"
說真ㄉ...我自認我"幾何"最爛= ="
不過我還是蠻喜歡作幾何題...
只要是數學~我什麼都好= =+
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Searchtruth - 訪客