Hello,
We suppose that b≠0 and d≠0.
[tex]1) \dfrac{a}{b} = \dfrac{c}{d}\ ==\textgreater ac=bd\\\\
2)\dfrac{a+b}{c+d} = \dfrac{a}{c}\\\\
Indeed \\\\
\dfrac{a}{b} = \dfrac{c}{d}\\ ==\textgreater\ ad=bc\\
==\textgreater\ ad+ac=bc+ac\\
==\textgreater\ a(c+d)=c(a+b)\\
==\textgreater\ \dfrac{a+b}{c+d}= \dfrac{a}{c}
[/tex]
[tex]\dfrac{ \sqrt{a^2+b^2}}{ \sqrt{c^2+d^2}} = \dfrac{a+b}{c+d} = \dfrac{a}{c}\\
==\textgreater\ \dfrac{a^2+b^2}{c^2+d^2} = \frac{a^2}{c^2}\\
==\textgreater\ a^2c^2+b^2c^2=a^2c^2+a^2d^2\\
==\textgreater\ (bc)^2-(ad)^2=0\\
==\textgreater\ (bc+ad) (bc-ad)=0\\
==\textgreater\ bc=-ad\ (exclude)\ or\ bc=ad
[/tex]
