Prove $\triangle$ AB$\Omega$ $\cong$ $\triangle$CD$\Lambda$. In omega triangle $\triangle AB\Omega$ and $\triangle CD\Lambda$, given $$\angle A \cong \angle B\,,~~\angle C\cong \angle D\,,~~\text{and}~~ \overline{AB} \cong \overline{CD}~,$$ prove $\triangle AB\Omega \cong  \triangle CD\Lambda$.
How do I start this I've attempted many ways with side angle and by angle angle the problem is the angles aren't congruent on different triangles they are congruent within their omega triangle. How would I go about this?
 A: Assume that in omega triangle $\triangle AB\Omega$ we have $\angle A \cong \angle B$.
Let $M$ be the midpoint of segment $\overline{AB}$, let $l$ be a line perpendicular to $\overleftrightarrow{AB}$ passing through $M$, and let $r$ be a ray emanating from $M$ contained in $l$ which lies on the same side of $\overleftrightarrow{AB}$ as rays $\overrightarrow{A\Omega}$ and $\overrightarrow{B\Omega}$.
We shall prove that rays $r$ and $\overrightarrow{A\Omega}$ are limiting parallel (by the same argument $r$ and $\overrightarrow{B\Omega}$ will be limiting parallel).
In the first step suppose that rays $r$ and $\overrightarrow{A\Omega}$ intersect at point $P$. Then triangles $\triangle AMP$ and $\triangle BMP$ are congruent and therefore $\angle AB\Omega=\angle B\cong \angle A=\angle BAP=\angle MAP\cong MBP=\angle ABP$ and by one of the axioms rays $\overrightarrow{MP}$ and $\overrightarrow{M\Omega}$ are equal. Hence $P$ is a point common to rays $\overrightarrow{AQ}$ and $\overrightarrow{BQ}$ which contradicts the assumption that these rays are parallel. We conclude that rays $r$ and $\overrightarrow{A\Omega}$ are disjoint.
Lemma. Let $\overrightarrow{A\Omega}$ and $\overrightarrow{B\Omega}$ be limiting parallel and a ray $r$ emanating from a point $M$ such that $A*M*B$ lie on the same side of $\overleftrightarrow{AB}$ as $\overrightarrow{A\Omega}$ and $\overrightarrow{B\Omega}$. If rays $r,\overrightarrow{A\Omega}$ and $r,\overrightarrow{B\Omega}$ are disjoint, then $r,\overrightarrow{A\Omega}$ and $r,\overrightarrow{B\Omega}$ are limiting parallel.
If you denote the midpoint od $\overline{CD}$ by $N$ you can easily prove by side-angle criterion that $\triangle AM\Omega\cong\triangle CN\Lambda$.
A: The conditions mean that both triangles are isosceles, and their bases are congruent.
Then you should see that they are congruent.
