Geometry Proof to Find Maximum area of $\triangle PIE$ 
Circle $\omega$ is inscribed in unit square $PLUM,$ and points $I$ and $E$ lie on $\omega$ such that
$U, I,$ and $E$ are collinear. Find, with proof, the greatest possible area for $\triangle PIE.$

I'm not sure if there is a solution possible without trigonometry.
Also, for my diagram in my solution, I'm not sure how to center it. Sorry about that.


Corners of unit square $U(0,0),P(1,1)\;$
Intersection between unit circle and straight line
$$ (x-\frac12)^2+(y-\frac12)^2= \frac14\;;y= \tan v\cdot x; \;$$
gives coordinates $(xI,yI), (xE, yE) $ for points $(I,E); \;$
Area (doubled)  is given by determinant
$$\big[(1,1,1),(xI,xE,1),(yI,yE,1)\big]$$
$$
t= \tan v \;;
\left(\text{xE}(\text{t$\_$})=\frac{t+\sqrt{2} \sqrt{t}+1}{2 \left(t^2+1\right)};\text{yE}(\text{t$\_$})=\frac{1}{2} \left(\frac{t^2}{t^2+1}+\frac{t}{t^2+1}+\frac{\sqrt{2} t^{3/2}}{t^2+1}\right);\right) \left(\text{xI}(\text{t$\_$})=\frac{t-\sqrt{2} \sqrt{t}+1}{2 \left(t^2+1\right)};\text{yI}(\text{t$\_$})=\frac{1}{2} \left(\frac{t^2}{t^2+1}+\frac{t}{t^2+1}-\frac{\sqrt{2} t^{3/2}}{t^2+1}\right);\right)
$$
Expressions of Area ( heavy, not pasted here ) and derivatives were handled by Mathematica . Graph of the derivative crosses x-axis at
$$\tan v \approx  0.267949 = \tan 15^{\circ}$$
Transversal that maximizes area of triangle $PIE$ is inclined at 15 degrees to x-axis. Area by substitution of critical  value is
$$ \Delta PIE=\dfrac14$$
which also agrees with result by Bongocat.

