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$\text{a)}x+y=c$  

Z twierdzenia Pitagorasa w trójkącie ABC:

$a^2+b^2={\left(x+y\right)}^2$  

$a^2+b^2=x^2+2xy+y^2$  

 

Z twierdzenia Pitagorasa w trójkątach ADC i BCD:

$\{\begin{array}{l}{x}^2+h^2=a^2\\ y^2+h^2=b^2\end{array}$  

A więc:

 

$a^2+b^2=x^2+2xy+y^2$  

$x^2+h^2+y^2+h^2=x^2+2xy+y^2$  

$2h^2=2xy$  

$h^2=xy$  

$h=\sqrt{xy}$  

$\left|CD\right|=\sqrt{\left|AD\right|\cdot \left|BD\right|}$  

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$\text{b)}{a}^2+b^2={\left(x+y\right)}^2$  

$a^2+b^2=x^2+2xy+y^2$  

 

Z podpunktu a wiemy, że:

$b^2=y^2+h^2$  

$h^2=xy$  

A więc:

 

$a^2+y^2+h^2=x^2+2xy+y^2$  

$a^2+h^2=x^2+2xy$  

$a^2+xy=x^2+2xy$  

$a^2=x^2+xy$  

$a^2=x\left(x+y\right)$  

$a=\sqrt{x\left(x+y\right)}$  

$\left|AC\right|=\sqrt{\left|AD\right|\cdot \left|AB\right|}$  

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$\text{c)}{a}^2+b^2=x^2+2xy+y^2$  

 

Z podpunktu a wiemy, że:

$a^2=x^2+h^2$  

$h^2=xy$  

A więc:

 

$a^2+b^2=x^2+2xy+y^2$  

$x^2+h^2+b^2=x^2+2xy+y^2$  

$h^2+b^2=2xy+y^2$  

$xy+b^2=2xy+y^2$  

$b^2=y^2+xy$  

$b^2=y\left(y+x\right)$  

$b=\sqrt{y\left(y+x\right)}$  

$\left|BC\right|=\sqrt{\left|BD\right|\cdot \left|AB\right|}$