论文部分内容阅读
由正弦定理出发,我们可以得到如下定理:△ABC中,以sinA、SinB、sinC为边可以构造△A′B′C′。且△ABC∽A′B′C′,△A′B′C′外接圆直径为1。证明:设△ABC外接圆半径为R, sinA+sinB=1/2R (a+b)>1/2R·C=sinC。同理可证 sinA+sinC>sinB,sinB+sinC>sinA。因此以sinA、sinB、sinC为边可以构造△A′B′C′。由正弦定理 a/sinA=b/sinB=c/sinC,因此△ABC∽△A′B′C′,则A=A′,B=B′,C=C′。设△A′B′C′外接圆半径为R′,对△A′B′C′施行正弦定理,则sinA/sinA′=2R′=1。由这个定理出发,有下面的二个应用。一、关于三角形中一些恒等式和不等式的互证
Starting from the sine theorem, we can obtain the following theorem: In ABC, △A’B’C’ can be constructed with sinA, SinB, and sinC as edges. And △ABC∽A′B′C′, ΔA′B′C′ has a circumscribed circle diameter of 1. Proof: Let ΔABC circumcircle radius R, sinA+sinB=1/2R (a+b)>1/2R·C=sinC. Similarly, it can be proved that sinA+sinC>sinB, sinB+sinC>sinA. Therefore, ΔA′B′C′ can be constructed with sinA, sinB, and sinC as edges. From the sine theorem a/sinA=b/sinB=c/sinC, thus ΔABC∽ΔA′B′C′, then A=A′, B=B′, and C=C′. Let ΔA′B′C′ have a radius of circumscribed circle R′ and apply sine theorem to ΔA′B′C′, then sinA/sinA′=2R′=1. Starting from this theorem, there are the following two applications. I. About mutual authentication of some identities and inequalities in a triangle