Respuesta :

Consider the figure,

From the right angled triangle ABD, using Pythagoras theorem, we have,

[tex]\begin{gathered} Hypote\nu se=\sqrt[]{base^2+height^2} \\ y=\sqrt[]{16^2+6^2} \\ \text{ =}17.08=17.1 \end{gathered}[/tex]

FRom the angle bisector theorem, we have,

[tex]\frac{AB}{BC}=\frac{AD}{CD}[/tex]

That is, we have,

[tex]\begin{gathered} \frac{16}{x}=\frac{17.1}{z} \\ \end{gathered}[/tex]

Thus, second option is correct.

Ver imagen LeianaT658322
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