SOLUTION: given: parallelogram ABCD, line DE is perpendicular to line AC, line BF is perpendicular to line AC prove: triangle ADE is congruent to triangle CBF

0. Make a sketch. It is intended not to replace the proof, but to help you follow the proof.

1. Consider these two triangles, ADE and CFB.
   Notice that these triangles are right-angled triangles.

2. Their sides AD and BC are congruent as the opposite sides of the parallelogram.

3. These triangles have congruent angles LDAE and LFCB, since these angles are alternate interior angles 
   at the parallel lines AD and BA and the transverse AC.

4. Since the right angled triangles ADE and CFB have the pair of congruent acute angles LDAE and LFCB, 
   they have the other pair of acute angles LADE and LCBF congruent too.

5. Now the triangles ADE and CFB are congruent according to the ASA-test of congruency of triangles.