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¡µABC congruent to ¡µBCA?(*)


·íµ§ªÌ­è¤J¦æ±Ð®Ñ®É¡A¤@¦¸©ó¤¤¥|¯ZùØ¡A¦b¶ÂªO¤W­pºâ¥k¹Ïªº¤@¹D¼ÆÃD¡G

¡ç A = ¡ç A (¦P¨¤)
¡ç ABP = ¡ç ACB (¤wª¾)
¡ç APB = ¡ç ABC (¤T¨¤§Î¤º¨¤©M)
Therefore, ¡µ ABP ~ ¡µ ABC (AAA)

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³o¤@Ãþ°ÝÃD¨ä¹ê¤]¬Æ´¶¹M¡C1991¦~¡A¦³¦ì¦Ñ®v¼g«Hµ¹­»´ä¼Æ²z±Ð¨|¾Ç·|(¨£1991¦~·|°T¡A·í®Éµ§ªÌ¥ç¦³¥h«H¦^À³)¡A°Ý3:4¬O§_µ¥©ó3/4¡C­n¦^µª³oÃþ°ÝÃD¡A¹ê¥i±q¤T­Ó¼h­±¥h¬Ý¡G

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¦]¦V¶q x1 , x2 , ... , xn ½u©Ê¬ÛÃö¡A¥²¦³¯Â¶q launda1, launda2,..., laundan«D¥þ¹s¨Ï±o

launda1 x1+ launda2 x2 +...+ laundan xn=0

¤£¥¢¤@¯ë©Ê¡A³] launda1 ¡Ú 0¡A¦³

x1=(- launda2/ launda1) +...+ laundan/ launda1) xn ¡C

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1-(1/2)+(1/3)-(1/4)+...+(-1)n-1(1/n)+... (-1¦¡-)¦)
=summation for r=1 to ¡Û (-1)r-1(1/r)(-2¦¡-)

= ln 2 (-3¦¡-)

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¡µABC congruent to ¡µBCA? - ­¶­º «e¤@³¹ «á¤@³¹ ¡m¼Æ¾Ç±Ð¨|¡n²Ä¤G´Á ¥Ø¿ý