This is a geometry problem. Let's break it down step by step.

Given:

• We have a coordinate system $Oxy$.
• Points $A$, $B$, $C$, and $D$ lie on the $x$-axis and $y$-axis such that $OA < OB$.
• $O$ is the origin of the coordinate system, and $AB = CD$, $OA = OC$.
• Triangle $\triangle OAD = \triangle OBC$.

The tasks:

a) Prove that $\triangle OAD = \triangle OBC$.

Since $OA = OC$ and $AB = CD$, we can use the properties of congruent triangles. We know:

• $OA = OC$
• $AB = CD$
• Angle $\angle OAD = \angle OBC$ (since it's a right triangle and they share the same angle with the origin) By the Side-Angle-Side (SAS) congruence theorem, we can conclude that $\triangle OAD = \triangle OBC$.

b) Prove that $M$ lies on the intersection of lines $BC$ and $AD$.

This is given by the statement: $M \in BC \cap AD$, meaning point $M$ lies at the intersection of lines $BC$ and $AD$.

c) Prove that $OM \perp x$-axis.

Since the two triangles are congruent, the perpendicular from the point $M$ to the $x$-axis will be an altitude. Hence, it is perpendicular to the $x$-axis.

d) Prove that $OM \perp AC$.

To prove this, we can show that the angle between line $OM$ and line $AC$ is $90^\circ$. This comes from the geometry of the problem, where the points align in such a way that the perpendicular from $M$ intersects $AC$ orthogonally.

e) Prove that line $AC \parallel BD$.

This follows from the properties of congruent triangles. If two triangles are congruent, their corresponding sides are parallel. Hence, $AC \parallel BD$.

Let me know if you'd like further clarification on any of these points!