Let's consider triangle ABC
Length AB can be obtained using Pythagoras
[tex]\begin{gathered} AB^2=x^2+y^2 \\ AB\text{ = }\sqrt[]{x^2+y^2} \\ \end{gathered}[/tex]
Similarly, we can consider triangle ACD, so that length AD will be obtained through Pythagoras
[tex]\begin{gathered} AD^2=x^2+z^2 \\ AD\text{ = }\sqrt[]{x^2+z^2} \end{gathered}[/tex]
Considering triangle ABD, with BD being the hypotenuse
[tex]\begin{gathered} BD^2=AD^2+AB^2 \\ (y+z)^{2\text{ }}=(x^2+z^2)+(x^2+y^2\text{)} \end{gathered}[/tex]
Expanding the parentheses
[tex]\begin{gathered} y^2+2yz+z^2=x^2+z^2+x^2+y^2 \\ \\ y^2-y^2+z^2-z^2+2yz=2x^2 \\ 2yz=2x^2 \\ \end{gathered}[/tex]
Divide both sides by 2
[tex]\begin{gathered} \frac{2yz}{2}=\text{ }\frac{2x^2}{2} \\ yz=x^2 \\ \\ x^2\text{ =yz} \\ \\ x\text{ = }\sqrt[]{yz} \end{gathered}[/tex]
Option A is correct
