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.