# 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?

• Look at Theorem 9.5 here – Quang Hoang Apr 23 at 17:07
• Quang Hoang thm 9.5 can’t be used because we aren’t given that corresponding angles are congruent – user597188 Apr 23 at 17:13
• Oh, that's true. Please ignore my comment – Quang Hoang Apr 23 at 17:17
• Please see MathJax to properly format math expressions. In particular, you should use only one pair of dollar signs to enclose an entire math expression, not on each symbol. – Lee David Chung Lin Apr 23 at 17:55
• Dr.Mathva wow....someone edited my post...I hate that this allows people to edit it without my approval first because that’s completely misleading...I’m sorry someone edited that as a tag I did not do that – user597188 Apr 23 at 19:00

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 $$\triangle AM\Omega\cong\triangle CN\Lambda$$.

The conditions mean that both triangles are isosceles, and their bases are congruent.

Then you should see that they are congruent.

• Being they are isosceles I made sketches using an ideal point and it shows they are always congruent is there a way to prove it using contradiction? Starting out by saying they aren’t congruent and proving they are – user597188 Apr 23 at 18:03