$$\frac{x}{y}=3$$

$$△AEH\approx △BFE$$

$$\frac{x}{y}=\frac{15-q}{p}=3$$

$$15-q=3·p$$

$$△AEH\approx △BFE$$

$$\frac{13-p}{q}=\frac{15-q}{p}$$

$$p·\left(13-p\right)=q·\left(15-q\right)$$

$$p·\left(13-p\right)=\left(15-3p\right)·\left[15-\left(15-3p\right)\right]$$

$$p·\left(13-p\right)=\left(15-3p\right)·\left(15-15+3p\right)$$

$$p·\left(13-p\right)=\left(15-3p\right)·3p$$

$$13p-p^2=45p-9p^2$$

$$9p^2-p^2+13p-45p=0$$

$$8p^2-32·p=0 |·\frac{1}{8}$$

$$p^2-4p=0$$

$$p\left(p-4\right)=0$$

$$q=15-3·p=15-3·4=15-12$$

Oba trougla su, pored toga što su slična, i pravougla. To nam omogućava da zapišemo Pitagorinu teoremu:

$$y^2=p^2+q^2$$

$$y^2=4^2+3^2$$

$$y^2=25$$

$$x=3·y=3·5$$