Challenging geometry problem: congruent triangles

AD is an altitude of triangle ABC. Points E, F and G are midpoints of the sides AB, BC and CA, respectively.
Prove that ∆DEG is congruent to ∆FGE
In ∆ABC, D is the midpoint of side AB. Extend BC to E so that CE = ½ BC. Draw DE, intersecting AC at F.
Prove that DF = FE

In ∆ABC with ∠ACB = 90°, CD is the altitude to hypotenuse AB. CM is the median to hypotenuse AB. CE is the bisector of ∠ACB.
Prove that CE is the bisector of ∠DCM.

BD and CE are altitudes to sides AC and AB of ∆ABC, respectively. F is the midpoint of BC, and G is the midpoint of DE.
Prove that FG _┴ DE.

In ∆ABC, BD and CE are bisectors of ∠ABC and ∠ACB. Perpendiculars drawn from A to BD and CE meet BD and CE at G and F respectively.
Prove that FG || BC.