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	<title>數學絲路</title>
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	<description>討論如何解數學問題與如何進行數學推論</description>
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		<title>我有一個夢</title>
		<link>http://witsofmath.wordpress.com/2008/10/23/ihaveadream/</link>
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		<pubDate>Thu, 23 Oct 2008 15:49:29 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[抒情]]></category>
		<category><![CDATA[教育]]></category>

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		<description><![CDATA[我有一個夢 我希望學校不再有煩人的段考與無窮無盡的小考（雖然我極贊成各種專業科目的檢定考），我希望學校是一個師生可以自由徜徉其間，討論共同感興趣的課題，思考人類過去、現在與未來所會面臨的問題（正如牛頓當年於蘋果樹下思考重力問題一般），並提出可行解決方案的地方，而不是讓學生只汲汲營營於分數，把學業當成獲取功名（考取高中、大學）的跳板，然而卻對自己的終身學習毫無助益的地方。我希望學生可以自由選擇自己真正想學的事物，而且學校也可以找到該領域學有專精的優秀人才來指導他們，而不是由一群自稱為教育專家或學校的老師來規定學生「應該」學什麼。 讓教導者可以指導有潛力的年輕人，讓年輕人可以學到他們想學的目標，讓他們可以快速的跟上前人、並超越前人，最重要的是：讓他們可以擁有許多的「空白」，讓他們有時間可以進行腦力激盪，甚至是體力的鍛鍊。而不是像目前的教育制度一樣，讓一大群的人匆匆忙忙到校（因為怕遲到）、渾渾噩噩於課堂間（因為一天要上七、八節，晚上還要補習），最後腦袋空空的回家，然後日復一日、年復一年，直到這個年輕人終於離開學校（如釋重負），變成一個不知人生目標、為何而活的無賴中年人為止，然後再去逼迫自己的下一代要「好好唸書」，以後才會「出人頭地」。 我不是一個害怕太多考試的中學生，我是一個教了十幾年數學的高中老師，看盡了太多浪費人生中最寶貴的青春歲月的高中生，他們悄悄地來，腦袋空空的走，不帶走一片雲彩。他們找不到自己想學的東西，校內的老師也對於學生的不專注而感到挫折，彼此浪費彼此的時間（我想這是上過學，尤其是已經上過「很多年學」的人，所共同擁有的記憶吧！）。 談過了一點點自己「卑微」的夢想，我想談一談目前的「教學現場」。因為我的領域是數學，所以我所談的主要是教導數學的經驗。 就我個人而言，數學能力的培養無非是下面幾個面向： 理解目前數學課程內容的能力（「主動學習」是這種能力的必要條件） 進行邏輯推理的能力（我認為這是中學數學老師最能著力的地方） 進行嚴密計算的能力（這是非數學專業的人認為學習數學的重點） 檢驗自己推理與計算結果的能力（學生通常忽略此點，甚至一般老師也很少如此做） 解決未知問題的能力（雖然這是我們學習的主要目標，但通常需要有一點「天份」的人才能真正達到此點） 但就我個人的經驗而言，「實況」是這樣，容我舉個例子： 「畢氏定理」與「相似三角形對應邊成比例」這樣的觀念，在數學來說，就像是英文裡的單字apple或book一樣基本，對吧？但隨便抓一個「高三的學生」來問問，知道為什麼「畢氏定理： 」的，寥寥可數；至於，知道為什麼「相似三角形對應邊成比例」的，就我個人目前的實測經驗是：一個也沒有！驚訝嗎？如果你不要抓一個「學生」，你去抓一個「中學數學老師」來問問：為什麼「相似三角形對應邊成比例」？恐怕實情更會令你驚訝，因為能馬上說出個為什麼的老師，恐怕也不多（不相信的人可以實驗看看！）。 有的學生甚至想用「正弦定理」來證明它，這根本是「本末倒置」型的推理錯誤，因為我們是先有「相似三角形對應邊成比例」的概念，然後才有「正弦函數」的出現，最後才導出「正弦定理」的，怎可用「正弦定理」來證明「相似三角形對應邊成比例」呢？（這個道理就好像是蓋房子一樣，你怎麼可能先蓋好三樓，然後再蓋一樓呢？） 為什麼會這樣？為什麼一個學生從小學、國中到高中，學了這麼多年的「數學」，結果卻連如此基本的定理如何來的都不知道？（這個現象倒有點像我們台灣學生，學了多年「英文」，結果到了大學畢業後，幾乎沒幾個人說得出口！） 我想，這跟我們的考試制度、社會的功利取向與老師、學生、家長一昧的求速成有很大的關係。我們的許多重要的數學考試（如大學學測等），只考選擇題，造成學生不重理解，只求「猜測準確」，老師無法得知學生是「做對」還是「猜對」。課程安排方面，也越來越不注重需要證明的內容，老師也樂得「跳過」這些吃力不討好的證明過程，反正學生也不一定聽得懂。再加上一直有不斷轟炸的大大小小考試，逼得學生們只好拼命的背誦「定理」、「公式」，忙於各種繁瑣的計算，而無暇顧及「理解、推理」才是學習數學的正道，以至於絕大部分的人學過中學數學、長大成人後，數學能力退化到小學程度（因為只會加減乘除！）。 這樣的過程（學了十多年數學，結果幾乎跟沒學是一樣的！），不只是浪費時間而已。關心教育的人們，我必須說：這是整個國家資源的重大浪費啊！學生虛擲了寶貴青春（我認為這是最重要的），老師消磨了教學熱忱，家長們也浪費了他們的金錢（當然包含補習費），整個國家（大部分）的數學教育呈現空轉狀態。 基於上述的種種原因，我認為：數學的教學現場是到了需要做些改變的時候了（雖然現在學生沒有選擇「不學」數學的權利，而身為老師的我們也沒有選擇「不教」的權利，所以難免還是會造成一些「怨偶」現象！）。 我們需要改變，因此我決定將採用傳統授課方式的教室（老師說學生聽，通常「不鼓勵」問問題，但反正台灣學生也已經被「訓練」成「沒什麼問題」！），改變成開放性的討論空間，同學、師生間可以進行任何形式的討論，主要以對談與問答的方式來腦力激盪，有時我會主動提出徵答問題，鼓勵學生深度思考，沒有固定課程（但遺憾的是，學校「當然」還是有固定的段考！），學生可以以自己的能力判斷，自行選擇高中課程的任何範圍，當作是自己學習的目標，我則擔任輔助的角色，在他有問題的時候（不管是課程內容的理解，還是解題遭遇困難），協助他渡過難關，但只要渡過這個難關，我也馬上放手，讓他繼續以自己的能力，回到他自行規劃的路線上。 我希望透過這樣的方式，經由適時引導（而不是喋喋不休的灌輸），讓學生能自然而然學會「邏輯推理」與「思想辯證」的過程，培養出「獨立判斷」與「自主學習」的精神，進而默默的達到「終身學習」的境界。我絕對鼓勵「自主學習」，我懷疑：如果凡事都需要老師叮嚀的學生，他能學到什麼？但，知我用心的學生（家長、甚或是教育界同仁），幾希？敢勇於突破傳統，不淪為「為考試而成為教育打手」的老師者，又幾希？ 多年前，一位就讀於知名美國大學的年輕人，由於學校教育無法滿足他的需求，所以他毅然輟學創業，多年後，造就了一個軟體帝國—「微軟」，他就是世界首富比爾蓋滋。稍晚，一位芬蘭大學生有感於微軟的獨佔事業過於龐大，所以決定開發「開放程式碼」的作業系統，於是令微軟不寒而慄的作業系統Linux應運而生。其他事業體也有感於微軟Office那舖天蓋地的佔有率不儘合乎公平正義，而開發出免費的線上文件系統，如Google文件、Zoho Docs等不勝枚舉。 當別的國家的人才們在考慮現今或未來的問題，並尋求解決之道的同時，關心台灣教育與國家發展的人們，您知道我們中學的學生們，我們未來的國家主人翁們在問什麼樣的問題嗎？ 「老師，這次段考考到哪裡？」 「老師，這題會考嗎？」 「老師，有沒有更快的方法？難道真的要那麼麻煩嗎？」 我想要問的問題是： 「教育同仁們，」 「我們到底應該怎樣做，才能讓我們的孩子們問『更大一點』的問題？」 一位希望為台灣教育盡點心力的數學老師 羅驥韡 留 Posted in 抒情<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=145&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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		<slash:comments>2</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
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	</item>
		<item>
		<title>指對數的親密接觸</title>
		<link>http://witsofmath.wordpress.com/2008/08/29/%e6%8c%87%e5%b0%8d%e6%95%b8%e7%9a%84%e8%a6%aa%e5%af%86%e6%8e%a5%e8%a7%b8/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/29/%e6%8c%87%e5%b0%8d%e6%95%b8%e7%9a%84%e8%a6%aa%e5%af%86%e6%8e%a5%e8%a7%b8/#comments</comments>
		<pubDate>Thu, 28 Aug 2008 16:18:38 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[函數]]></category>
		<category><![CDATA[對數]]></category>

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		<description><![CDATA[此篇文章會牽涉到一些指數的「微分」，所以如果你要看懂全文的話，你可能需要一點「微積分」的知識。 學過高中數學的人會知道：「指數函數 與對數函數 會對稱 45° 線」。 不管底數 a (a &#62; 0 , a ≠ 1) 怎麼變化，兩者的圖形始終會保持對稱。 參考圖檔：GeoGebra 圖 現在問題來了： 什麼樣的 a 會讓兩者相切（如右圖）？ 它們相切的時候，切點在哪裡？     解答： 當它們交會在 這條直線上時，因為交點座標  位於  上，而且此點的切線斜率為 1，所以此交點會符合兩個方程式： 的微分（也就是 的切線斜率）為 稱為「自然對數」，也就是以「尤拉數」e ≒ 2.718281 為底的對數函數 將第 式代入第 式，我們可以得到： 所以： 代回第 式，可得： 兩邊取「自然對數」，可得： 然後，兩邊取「自然指數」，可得： 再取一次「自然指數」，可得： 將此結果代回第 式，可得： 從以上的討論，我們得到一些結論： 切點座標為  指數函數和對數函數的底數等於  時，它們才會剛好相切。 「尤拉數」e ≒ 2.718281 [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=135&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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		<slash:comments>4</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
		</media:content>

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		<media:content url="http://witsofmath.files.wordpress.com/2008/08/log_exp2-custom.png" medium="image">
			<media:title type="html">指對數第二次接觸</media:title>
		</media:content>

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			<media:title type="html">指對數交於三點</media:title>
		</media:content>
	</item>
		<item>
		<title>常用的圖檔</title>
		<link>http://witsofmath.wordpress.com/2008/08/28/%e5%b8%b8%e7%94%a8%e7%9a%84%e5%9c%96%e6%aa%94/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/28/%e5%b8%b8%e7%94%a8%e7%9a%84%e5%9c%96%e6%aa%94/#comments</comments>
		<pubDate>Wed, 27 Aug 2008 19:20:45 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[管理用]]></category>

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		<description><![CDATA[小圖： &#160; item 1 item 2 方程式 分隔線：<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=128&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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		<slash:comments>0</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
		</media:content>

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	</item>
		<item>
		<title>log3x 尾數的兩倍</title>
		<link>http://witsofmath.wordpress.com/2008/08/26/log3x-%e5%b0%be%e6%95%b8%e7%9a%84%e5%85%a9%e5%80%8d/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/26/log3x-%e5%b0%be%e6%95%b8%e7%9a%84%e5%85%a9%e5%80%8d/#comments</comments>
		<pubDate>Mon, 25 Aug 2008 22:45:51 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[解題]]></category>
		<category><![CDATA[代數]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/08/26/logx-and-log3x/</guid>
		<description><![CDATA[log(x) 的尾數是 log(3x) 的尾數的 2 倍，有這樣的 x 嗎（x為正實數）？ 解說： 當然，在瞭解這個問題之前，你必須先知道什麼是對數中的「首數」與「尾數」。 我先舉一些例子來說明這個問題的特性。 首先，假設 x 落在下面的區間中（如右圖）： 則： 因此，我們可以知道： （因為 x 在 10 到 100 之間） （因為 3x 在 10 到 100 之間） 所以： log(x) 的首數 = 1，尾數 = log(x) &#8211; 1；log(3x) 的首數 = 1，尾數 = log(3x) &#8211; 1 因此如果這個區間內有任何 x 會吻合「log(x) 的尾數是 log(3x) 的尾數的 2 倍」這個條件的話，那麼 x [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=117&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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			<media:title type="html">羅驥韡</media:title>
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	</item>
		<item>
		<title>「首數」與「尾數」</title>
		<link>http://witsofmath.wordpress.com/2008/08/25/mantissa/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/25/mantissa/#comments</comments>
		<pubDate>Sun, 24 Aug 2008 21:41:51 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[代數]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/08/25/mantissa/</guid>
		<description><![CDATA[到底什麼是對數中的「首數」與「尾數」呢？ 首先，我們先舉幾個例子來說明。 例一： 當 log(x) = 3.265 時，log(x) 的「首數」就是 3.265 的整數部份，也就是 3 log(x) 的「尾數」就是 3.265 的小數部份，也就是 0.265 例二： 當 log(x) = -3.265 時，log(x) 的「首數」就是在 -3.265 *左邊的整數*，也就是 -4 log(x) 的「尾數」就是 -4 到 -3.265 的小數部份，也就是 0.735 從上面的兩個例子看來，「首數」和「尾數」的意思好像不一樣，其實不然。不管是「例一」或「例二」，log(x) 的首數都是指位於 log(x) 本身左邊的第一個整數（當然，如果 log(x) 本身就是整數的話，log(x) 的首數就是它本身），換句話說，我們可以用「高斯符號」來表示「首數」： 定義 log(x) 的首數 = [log(x)] log(x) 的尾數 = log(x) &#8211; [log(x)] 注意： 尾數 &#60; [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=106&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
		<wfw:commentRss>http://witsofmath.wordpress.com/2008/08/25/mantissa/feed/</wfw:commentRss>
		<slash:comments>1</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
		</media:content>

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	</item>
		<item>
		<title>中線長公式</title>
		<link>http://witsofmath.wordpress.com/2008/08/21/medianlength/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/21/medianlength/#comments</comments>
		<pubDate>Wed, 20 Aug 2008 17:18:43 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[幾何]]></category>
		<category><![CDATA[平行四邊形]]></category>
		<category><![CDATA[平方和]]></category>
		<category><![CDATA[旋轉]]></category>
		<category><![CDATA[中線]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/08/21/medianlength/</guid>
		<description><![CDATA[三角形「中線長」公式：右圖中， x 為三角形 c 邊上的中線，則： 證明： 首先，我們將整個三角形沿著 c 邊的中點旋轉 180°，會造成如右圖的平行四邊形，因此，我們可以進一步根據「平行四邊形定理」（平行四邊形中，兩對角線的平方和等於四邊的平方和）來計算中線長 x： 整理後可以得到： 兩邊開根號後，我們就可以得到：<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=86&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
		<wfw:commentRss>http://witsofmath.wordpress.com/2008/08/21/medianlength/feed/</wfw:commentRss>
		<slash:comments>0</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
		</media:content>

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	</item>
		<item>
		<title>二項分配圖</title>
		<link>http://witsofmath.wordpress.com/2008/08/15/binomialdistribution/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/15/binomialdistribution/#comments</comments>
		<pubDate>Fri, 15 Aug 2008 01:13:21 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[統計]]></category>
		<category><![CDATA[機率]]></category>
		<category><![CDATA[試算表]]></category>
		<category><![CDATA[二項分配]]></category>
		<category><![CDATA[二項式定理]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/?p=60</guid>
		<description><![CDATA[如果連續丟一個硬幣 32 次，那麼共出現 5 次正面的機率是多少？共出現 10 次正面的機率是多少？共出現 25 次正面的機率是多少？通常這樣的機率問題，需要利用「二項式定理」來計算。 如果你把正面出現的次數的所有可能性（可能是 0 次到 32 次）的機率全部計算出來的話，那麼你就可以將這些數據畫成一個稱為「二項分配圖」的統計圖表（如圖）。 試算表網址：http://www.editgrid.com/user/pegasusroe/二項分配 如果你進入上面的試算表網址中，丟硬幣的總次數是固定的 ( n = 32 )，但是硬幣出現正面的機率 p （預設 p = 0.5）和 q 是可以更改的，您可以更改 p、q 的數值看看二項分配圖會如何跟著改變，你也可以觀察不同的「出現正面的機率」之間，它們的分配圖有何異同，但請注意：因為 p、q 是機率值，所以只能介於 0 與 1 之間喔！<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=60&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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		<slash:comments>0</slash:comments>
	
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			<media:title type="html">羅驥韡</media:title>
		</media:content>

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			<media:title type="html">二項分佈圖（2）</media:title>
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		<item>
		<title>梯形中線</title>
		<link>http://witsofmath.wordpress.com/2008/08/12/trapezoidmedian/</link>
		<comments>http://witsofmath.wordpress.com/2008/08/12/trapezoidmedian/#comments</comments>
		<pubDate>Tue, 12 Aug 2008 00:01:42 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[解題]]></category>
		<category><![CDATA[幾何]]></category>
		<category><![CDATA[相似]]></category>
		<category><![CDATA[平行四邊形]]></category>
		<category><![CDATA[梯形]]></category>
		<category><![CDATA[中線]]></category>
		<category><![CDATA[中線長]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/08/12/trapezoidmedian/</guid>
		<description><![CDATA[如右圖，梯形 ABCD 的邊長分別為 12, 21, 30, 15。假設 x 為 AD 中點到 BC 中點的距離，請問 x = ? 解法一： 如果我們從「上底」畫兩條垂直線段下來（如右圖），並且把「下底」劃分為三部份（假設 12 右邊那一小段長為 a，且設高為 b），那麼我們就可以得到兩個高一樣的直角三角形，因此我們可以利用「畢氏定理」得到下面的關係式： 兩式相減可得： 因此，我們知道 a = 3，代入原來的方程式可得： 所以 我們把原來的中點連線畫回來後，會發現 x, 6,  三邊會形成直角三角形，所以我們又可以利用「畢氏定理」來計算 x： 因此， 解法二： 這一次我們把 AB 和 CD 延長，並假設它們交於 E 點，事實上，將中點連線 MN 延長也會通過 E 點（請讀者想一想為甚麼？） 如右圖，我們繼續作 AF 平行 CD，AG 平行 MN。因為 與 相似，EN 為 [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=52&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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			<media:title type="html">羅驥韡</media:title>
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		<item>
		<title>正三角形內的三線段</title>
		<link>http://witsofmath.wordpress.com/2008/07/18/20080718/</link>
		<comments>http://witsofmath.wordpress.com/2008/07/18/20080718/#comments</comments>
		<pubDate>Fri, 18 Jul 2008 06:01:31 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[解題]]></category>
		<category><![CDATA[幾何]]></category>
		<category><![CDATA[面積]]></category>
		<category><![CDATA[分割]]></category>
		<category><![CDATA[旋轉]]></category>
		<category><![CDATA[正三角形]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/07/18/20080718/</guid>
		<description><![CDATA[已知正三角形內一點到三頂點的距離分別為 6,8,10，求正三角形面積？ 附檔： GeoGebra 圖檔 解法： 首先，我們將圖中紅色的頂點固定，然後將原來的正三角形順時針旋轉 60° 接著，我們分別固定另外兩個頂點，然後同樣的將原來的正三角形順時針旋轉 60°，這樣我們就可以得到三個跟原來一模一樣的正三角形。 請觀察右圖中塗色的部份，它剛好是原來三角形面積的 2 倍。 現在我們觀察「同一塊面積」，但用不同的切割方式。在右圖中，你會發現上圖中的塗色部份，現在可以看成邊長分別為 6, 8, 10 的「正三角形」與三個邊長為 6, 8, 10 的三角形的面積總和。 綜合以上推論，我們可以得到： (ΔABC 面積)﹢2 = 最後，我們推得： ΔABC 面積 ＝<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=47&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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			<media:title type="html">羅驥韡</media:title>
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		<item>
		<title>圓內接四邊形裡面的圓</title>
		<link>http://witsofmath.wordpress.com/2008/07/17/20080711/</link>
		<comments>http://witsofmath.wordpress.com/2008/07/17/20080711/#comments</comments>
		<pubDate>Thu, 17 Jul 2008 01:37:25 +0000</pubDate>
		<dc:creator>羅驥韡</dc:creator>
				<category><![CDATA[解題]]></category>
		<category><![CDATA[幾何]]></category>
		<category><![CDATA[圓]]></category>
		<category><![CDATA[四邊形]]></category>
		<category><![CDATA[相似形]]></category>

		<guid isPermaLink="false">http://witsofmath.wordpress.com/2008/07/17/20080711/</guid>
		<description><![CDATA[如右圖，已知： ABCD 為圓內接四邊形 對角線 AC 與 BD 交於 P 點 M 為 BC 中點 虛線圓通過 P 點，切於 M 點，並與 AC、BD 交於 R、Q 兩點 線段 DS 與 BQ 等長 TS 平行 AD 試證明：線段 AT 與 CR 等長 附檔：GeoGebra 圖檔 [證明] 根據圓的「圓冪性質」，我們知道： 但因為 M 為中點，因此 ，所以： 我們重新將上面的等式整理成「比例」的形式，可得： &#8230;&#8230;.. 另外，從圖中我們可以看到：ΔPBC 與 ΔPDA 相似，所以： &#8230;&#8230;.. 最後，因為 TS 平行 AD，所以我們有以下的比例關係： [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=witsofmath.wordpress.com&amp;blog=3932227&amp;post=41&amp;subd=witsofmath&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
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			<media:title type="html">羅驥韡</media:title>
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