Cevian in Isosceles Right Triangle

Stewart's theorem states (with relation to the given triangle) that

$(\overline{AB})^2 \overline{PC} + (\overline{AC})^2 \overline{PB} = \overline{BC} ((\overline{PB})(\overline{PC}) + (\overline{PA})^2)$

Given the quantities $\overline{PB} = 3$ and $\overline{PC} = 11$ , we have that $\overline{BC} = 14$ and since the triangle $\overline{ABC}$ is an isosceles right triangle, we also have that $\overline{AB} = \overline{AC} = \frac{14}{\sqrt{2}}$ . Substituting the required quantities into the statement of Stewart's theorem, we have the following:

$14\left(\frac{14}{\sqrt{2}}\right)^2 = 14 (33 + (\overline{PA})^2)$

Dividing both sides by $14$ and computing that $\left(\frac{14}{\sqrt{2}}\right)^2 = \frac{196}{2} = 98$ , we have that

$(\overline{PA})^2 = 65,$

meaninig that $\overline{PA} = \sqrt{65}$ . Round off the answer as you may (that's if you believe in any of the square root jargon).

P.S. I have looked at a reformulation of Stewart's theorem using only the framework of Rational trigonometry; view my video and Wildberger's at the following links given.