Question

斯坦纳-雷米欧司定理的图不是反证法的图 是证明一：设三角形ABC,角B、角C的平分线是BE、CD 作∠BEF=∠BCD;

Answer 1

雷米欧司1840年写信给斯坦纳时提出了这个问题,下面这种证明是1842年就出现的,速度惊人啊,对于这样一个难题来说.

A proof using SSA:F.G.Hesse (1842)

C

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F___________E_____D

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/ G

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A/_______________B

We consider given that ABC is a triangle,

AD and BE bisect angles A and B such that AD=BE.

Construct F such that AF=AE and DF=AB.

Let G be the intersection of AD and BE.

We find ADF and EBA are congruent,so