Problem
GEO-B1-M07-P017 Parallelism From Congruent Triangles
#17
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(M\) is the midpoint of \(BC\). Extend \(AM\) beyond \(M\) to \(D\), where \(MD=AM\). Prove that \(AB\parallel CD\) through triangle congruence.
Prove \(\triangle ABM\cong\triangle DCM\), then compare angles.
We have \(AM=MD\), \(BM=MC\), and \(\angle AMB=\angle DMC\) as vertical angles. Thus \(\triangle ABM\cong\triangle DCM\). Therefore \(\angle ABM=\angle DCM\). These are alternate interior angles for lines \(AB\) and \(CD\) with transversal \(BC\), so \(AB\parallel CD\).
The same construction, but focused on hidden congruent triangles rather than the parallelogram criterion.