imc

Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below. Each of the angles at $X$ is $\frac{180^{\circ }}{9}=20^{\circ }$. From $△XYZ$, the degree measure of the smaller interior angle of the trapezoid is $\frac{180^{\circ }-20^{\circ }}{2}=80^{\circ }$, hence the degree measure of the larger interior angle is $180^{\circ }-80^{\circ }=\boxed{100}$. Proof that all the extended trapezoid legs intersect at the same point: It is sufficient to prove this for any pair of neighboring trapezoids. For two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg at the same point, Q.E.D. Knowing this, we can now easily see that the intersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. And by symmetry the point must be in the center of this side.