Załóżmy, że [tex]m_1=\frac{a+b}{a-b}, \ \ \ m_2=\frac{c+d}{c-d}, \ \ \ m_3=\frac{ac-bd}{ad+bc}[/tex].
Wtedy
[tex]m_1+m_2+m_3= \frac{a+b}{a-b} + \frac{c+d}{c-d} + \frac{ac-bd}{ad+bc}= \\ \\ = \frac{(a+b)(c-d)(ad+bc)}{(a-b)(c-d)(ad+bc)} + \frac{(c+d)(a-b)(ad+bc)}{(c-d)(a-b)(ad+bc)} + \frac{(ac-bd)(a-b)(c-d)}{(ad+bc)(a-b)(c-d)}= \\ \\ = \frac{(a+b)(c-d)(ad+bc)+(c+d)(a-b)(ad+bc)+(ac-bd)(a-b)(c-d)}{(a-b)(c-d)(ad+bc)}=\\ \\ = \frac{(ac-ad+bc-bd)(ad+bc)+(ac-bc+ad-bd)(ad+bc)+(ac-bd)(a-b)(c-d)}{(a-b)(c-d)(ad+bc)}= \\ \\ =\frac{(ad+bc)(ac-ad+bc-bd+ac-bc+ad-bd)+(ac-bd)(a-b)(c-d)}{(a-b)(c-d)(ad+bc)}=[/tex]
[tex]=\frac{2(ad+bc)(ac-bd)+(ac-bd)(ac-ad-bc+bd)}{(a-b)(c-d)(ad+bc)} = \frac{(ac-bd)(2ad+2bc+ac-ad-bc+bd)}{(a-b)(c-d)(ad+bc)}=\\ \\ = \frac{(ac-bd)(ad+bc+ac+bd)}{(a-b)(c-d)(ad+bc)}=\frac{(ac-bd)[d(a+b)+c(b+a)]}{(a-b)(c-d)(ad+bc)}=\\ \\ =\frac{(ac-bd)(a+b)(c+d)}{(a-b)(c-d)(ad+bc)}=\frac{a+b}{a-b}\cdot\frac{c+d}{c-d}\cdot\frac{ac-bd}{ad+bc} =m_1\cdot m_2\cdot m_3[/tex]
Zatem wykazaliśmy zadaną równość.
