Two Right Triangles Sharing a Hypotenuse

Two right triangles $ABC$ and $ADC$ share the common hypotenuse $AC = 10$ and lie on opposite sides of it, so that $\angle ABC = \angle ADC = 90^\circ$, forming the quadrilateral $ABCD$ shown. It is given that $\angle BAC = \dfrac{\pi}{3}$ and $\angle DAC = \dfrac{\pi}{4}$. The segment $BD$ meets $AC$ at the point $E$.

1. Find $BD^2$. 2. Find $AE$.