Do you remember this theorem?

Relevant wiki: Power of a Point

By the power of a point , we have

$(AP)^2=(AB)(AC)$

Substituting, we get

$(3\sqrt{2})^2=(AB)(AC)$

However, $AB=BC$ , therefore, $AC=2AB$ , so

$9(2)=(AB)(2AB)$

$18=2(AB)^2$

$9=(AB)^2$

$3=AB$

Finally,

$AC=2AB=2(3)=$ $\boxed{6}$

Let be $AB=x$ and $AC=2x$

By power of a point

$AP^2=AB\cdot AC$

$(3\sqrt{2})^2=2x\cdot x\Rightarrow x^2=9 \Rightarrow x=3$

$\therefore \boxed{AC=6}$

By the power of a point,

$(PA)^2=(AB)(AC)$

$(3\sqrt{2})^2=AB(2AB)$

$AB=3$

Therefore, $AC=6$ .

This can be found on Euclid's Elements: Book 3: Proposition 36.

By the power of a point, we have

$(AP)^2=(AC)(AB)$

$(3\sqrt{2})^2=(AB+AB)(AB)$

$18=2(AB)^2$

$9=(AB)^2$

$3=AB$

It follows that $AC=3+3=\boxed{6}$

We know by Power of a Point that (in this case): $AC\cdot AB=AP^2$ . If $AP=3\sqrt{2}$ , then $AP^2=18$ . We now know that $AC\cdot AB=18$ . Let's set length $AB=y$ and $AC=2y$ since $AB=BC$ Now we can form an equation and solve for $y$ :

$2y^2=18$

$y^2=9$

$y=3$ .

We know that $y=3$ works, so multiplying $AC\cdot AB$ gives us $6\cdot 3=18$ , so length $AC=6$