正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4

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正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4正七边形A1A2A3A4A5A6

正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4
正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4

正七边形A1A2A3A4A5A6A7,证明:1/A1A2=1/A1A3+1/A1A4
正七边形边长为a,一条对角线长为b,另一长度不等的对角线为c,
求证:1/a=1/b + 1/c
证明：(请画图,比较容易看清下面内容)设正七边形ABCDEFG中AB=BC=CD=a ,AC=b ,AD=c
延长AB、DC交于P,
则ΔPAD为等腰三角形,
设PB=PC=X
因为BC‖AD ,所以PC/PD = BC/AD ,即X/(x+a) = a/c
在ΔPAC中 ∠CAP=180/7 度,∠ACP=∠ADC+∠DAC=360/7 +180/7=540/7 度
所以∠P =180 -∠ACP-∠CAP=540/7 度
所以∠P=∠ACP ,
所以AP=AC ,即a+x=b
把x=b-a 代入X/(x+a) = a/c 中,
化简即为
1/a=1/b + 1/c

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