As $AB || CD$, we get $\angle EBA = \angle EDC$. Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$. So, $\angle EBA = \angle ABM$. Similarly we get $\angle EAB = \angle BAM$. So,we get $\triangle EBA \cong \triangle ABM$ by the ASA theorem. Then, we get $EM \perp AB.$ Now, let the segmen...