Angle in Square

The following problem has been posted by Miguel Ochoa Sanchez at the CutTheKnotMath facebook page. Miguel and Daniel Hardisky have posted practically identical solutions. The one below is due to Daniel.

Note that a converse problem is considered elsewhere.

Assume the side of the square equals $1.$ Since $\angle AEF=\angle APB$ and triangles $APQ$ and $AFE$ share the angle at $A$ they are similar. It is also given that their areas are in the ratio $1:2,$ $[\Delta AFE]/[\Delta APQ]=2.$

Drop the altitudes $AM$ and $AN$ and note that $AM^{2}/AN^{2}=[\Delta AFE]/[\Delta APQ]=2$ and that $AN$ is half the diagonal $AC$ of the square so that $AN=\sqrt{2}/2.$ From this it follows that $AM=1,$ implying that, as right triangles, $\Delta ADF=\Delta AMF$ and $\Delta ABE=\Delta AME.$ It follows that $\angle BAE=\angle MAE$ and $\angle DAF = \angle MAF$ so that

$\angle BAD = 2(\angle MAE +\angle MAF) = 2\angle EAF.$

Thus, indeed, $\angle EAF = 45^{\circ}.$

The configuration has additional properties. Two can be shown by simple angle chasing.

Indeed, $\angle AEB=90^{\circ}-\beta=\angle AEM=\angle AQP,$ implying $Q\in (ABE),$ a circle with diameter $AE.$ Thus $\angle AQE=90^{\circ}.$ Similarly, $FP\perp AE.$

Indeed, $\angle EQF=90^{\circ},$ making $EF$ a diameter of $(PEFQ).$ Since, $\angle ECF=90^{\circ},$ $C\in (PEFQ).$

More accurately,