So DE, DF and DF are the center lines of triangle AABC respectively.

So AE=CE

DE= 1/2BC

Parallel BC

So angle DEQ= angle EMC

DF parallel AC

DF= 1/2AC

So the quadrilateral DECF is a parallelogram.

So angle C= angle EDF

Angle DFP= angle c

Because AC=BC

So DE=DF

Because angle PDF= angle PDQ+ angle QDF

Angle QDE= angle QDF+ angle EDF

Angle PDQ= angle c

So angle PDF= angle QDE

Because DQ=DP

So triangle PDF and triangle QDE are congruent (SAS).

So the angle DFP= the angle DEQ.

So angle EMC= angle C.

So ME=CE

So AE = me

So Angel May = Angel MAE

Because angle AME+ angle MAE+ angle EMC+ angle C= 180 degrees.

Therefore, angle AME+ angle EMC= angle AMC=90 degrees.

So AM is perpendicular to BC

(2) Prove: Because points D, E and F are the midpoints of three sides of triangle ABC respectively.

So de and df are the center lines of triangle ABC respectively.

So AE=CE

DE= 1/2BC

Parallel BC

So angle DEM= angle EMC

DF= 1/2AC

DF parallel AC

So angle DFM= angle C.

Quadrilateral DECF is a parallelogram.

So DF=CE

Angle C= angle EDF

Because angle PDQ= angle c

Angle PDQ= angle PDE+ angle EDQ

Angle EDF= Angle PDE+ Angle PDF

So angle EDQ= angle PDF

Because AB=BC

So DE=DF

Because DQ=DP

So triangle EDQ and triangle FDP are congruent (SAS).

So angle DEQ= angle DFP

Because the angle DEQ+ the angle DEM= 180 degrees.

Angle DFP+ Angle DFM= 180 degrees.

So angle DEM= angle DFM

So angle EMC= angle C.

So ME=CE

So DF = me

(3)DF equals me.