Question - Analysis of Geometric Similarity and Area Ratio in a Quadrilateral

\angle AXM = \angle CXD \quad (\text{vertically opposite angles are equal})

\angle AMX = \angle CDX \quad (\text{corresponding angles of parallel lines are equal})

\angle A = \angle C \quad (\text{given})

\text{By AA similarity criterion,} \triangle AXM \sim \triangle CXD.

\frac{\text{area of }\triangle AMX}{\text{area of }\triangle CXD} = \left(\frac{AM}{CD}\right)^2

\text{Since }\triangle AXM \sim \triangle CXD\text{, their sides are proportional. Therefore, } \frac{AM}{CD} = \frac{AX}{CX}

\text{Therefore, the area ratio is } \left(\frac{AX}{CX}\right)^2.