12.69. Find the height of a trapezoid with bases $A B$ and $C D$ equal to $a$ and $b(a<b)$, the angle between the diagonals is $90^{\circ}$, and the angle between the extensions of the lateral sides is $45^{\circ}$.
12.69. Complete the triangle $ABC$ to form a parallelogram $ABCE$ (Fig. 12.8). Let $BC = x$ and $AD = y$. Then $(b-a) h = 2 S_{AED} = x y \sin 45^{\circ}$ and $(b-a)^{2} = x^{2} + y^{2} - 2 x y \cos 45^{\circ} = x^{2} + y^{2} - 2 x y \sin 45^{\circ}$. By the Pythagorean theorem, $a^{2} + b^{2} = (AO^{2} + BO^{2}) + (CO^{2} + DO^{2}) = (BO^{2} + CO^{2}) + (DO^{2} + AO^{2}) = x^{2} + y^{2}$. Therefore, $(b-a)^{2} = x^{2} + y^{2} - 2 x y \sin 45^{\circ} = a^{2} + b^{2} - 2(b-a) h$, i.e., $h = ab / (b-a)$.