Does a parellelogram have four congruent triangles?

Answer 1

Yes. Read on for why:

Take a parallelogram ABCD with midpoints E and F in the bases. So something like this (forgive the "drawing"):

A E B

__.__

/__.__/

C F D

We know that parallelogram AEFC = EBDF, since they have the same base (F bisects CD, so CF = FD), height (haven't touched that), and angles (<ACF = <EFD because they're parallel - trust me that everything else matches).

We also know that every parallelogram can be divided into two congruent triangles along their diagonal. So if two congruent parallelograms consistent of two congruent triangles each, then all four triangles are congruent.

So your congruent triangles are ACF, AEF, EFD, and EBD.

You can further reinforce this through ASA triangle congruency proofs (as I did at first), but this is a far more concise and equally valid answer.