a) 

Z rysunków wynika , że trójkąty ABC i A'B'C' są równoboczne : 

$\left|A{}^{\prime }C{}^{\prime }\right|=\left|AC\right|=4$  

$\left|AB\right|=\left|A{}^{\prime }B{}^{\prime }\right|=4$

$\left|BC\right|=\left|B{}^{\prime }C{}^{\prime }\right|=4$

 

$\left|\angle BAC\right|=\left|\angle B{}^{\prime }A{}^{\prime }C{}^{\prime }\right|=$   ${60}^o$  

$\left|\angle A{}^{\prime }B{}^{\prime }C{}^{\prime }\right|=\left|\angle ABC\right|=$   ${60}^o$

$\left|\angle ACB\right|=\left|\angle A{}^{\prime }C{}^{\prime }B{}^{\prime }\right|=$ ${180}^o-2\bullet {60}^o={180}^o-{120}^o={60}^o$ ${180}^o-2\bullet {60}^o={180}^o-{120}^o={60}^o$  

b) 

$\left|A{}^{\prime }F{}^{\prime }\right|=\left|AF\right|=3$

$\left|AB\right|=\left|A{}^{\prime }B{}^{\prime }\right|=5$

$\left|B{}^{\prime }C{}^{\prime }\right|=\left|BC\right|=5$    

$\left|CD\right|=\left|C{}^{\prime }D{}^{\prime }\right|=2$  

$\left|DE\right|=\left|D{}^{\prime }E{}^{\prime }\right|=\left|BC\right|-\left|AF\right|=5-3=2$  

$\left|EF\right|=\left|E{}^{\prime }F{}^{\prime }\right|=\left|AB\right|-\left|CD\right|=5-2=3$  

 

$\left|\angle F{}^{\prime }A{}^{\prime }B{}^{\prime }\right|=\left|\angle FAB\right|=$ ${90}^o$  

$\left|\angle A{}^{\prime }B{}^{\prime }C{}^{\prime }\right|=\left|ABC\right|=$   ${90}^o$

$\left|\angle BCD\right|=\left|\angle B{}^{\prime }C{}^{\prime }D{}^{\prime }\right|=$   $90{º}^o$

$\left|\angle C{}^{\prime }D{}^{\prime }E{}^{\prime }\right|=\left|\angle CDE\right|=$   $90º$    

$\left|\angle DEF\right|=\left|\angle D{}^{\prime }E{}^{\prime }F{}^{\prime }\right|=$   ${360}^o-{90}^o={270}^o$  

$\left|\angle EFA\right|=\left|\angle E{}^{\prime }F{}^{\prime }A{}^{\prime }\right|=$   ${90}^o$    

c) 

Figura KLMNO składa się z kwadratu KNOL i trójkąta równobocznego LNM : 

$\left|K{}^{\prime }L{}^{\prime }\right|=\left|KL\right|=3$  

$\left|L{}^{\prime }M{}^{\prime }\right|=\left|LM\right|=3$

$\left|MN\right|=\left|M{}^{\prime }N{}^{\prime }\right|=3$   

$\left|NO\right|=\left|N{}^{\prime }O{}^{\prime }\right|=3$  

$\left|KO\right|=\left|K{}^{\prime }O{}^{\prime }\right|=3$

 

$\left|\angle O{}^{\prime }K{}^{\prime }L{}^{\prime }\right|=\left|\angle OKL\right|=$   ${90}^o$

$\left|\angle KLM\right|=\left|\angle K{}^{\prime }L{}^{\prime }M{}^{\prime }\right|=$   ${90}^o+{60}^o={150}^o$  

$\left|\angle LMN\right|=\left|\angle L{}^{\prime }M{}^{\prime }N{}^{\prime }\right|=$ ${60}^o$  

$\left|\angle MNO\right|=\left|\angle M{}^{\prime }N{}^{\prime }O{}^{\prime }\right|=$   ${90}^o+{60}^o={150}^o$

$\left|\angle NOK\right|=\left|\angle N{}^{\prime }O{}^{\prime }K{}^{\prime }\right|=$   ${90}^o$