Then CEQB is a parallelogram, BQ=CE=CG, BC=CD, ∠QBC= 180-∠ECB=∠GCD.
So: △ CBQ △ GCD
So: ∠QCB=∠CDG, QC=GD.
Because: ∠ CDG+∠ MCD = ∠ QCB+∠ MCD =180-∠ BCD = 90.
So: CM is perpendicular to GD
The following proves that m and I coincide:
Because: ∠QCD=∠QCB+∠BCD=∠CDG+∠ADC=∠ADG.
AD=CD,QC=GD
So: △ QDC △ ADG
Because: advertising vertical DC
So: △QDC is equivalent to △ADG rotating 90 clockwise and translating with point C as the origin.
So: AG vertical QD
Similarly: △ FGD △ QCG
FD vertical QG
Therefore: QM, AG, DP*** Point P, (P is △QGD).
So: M and I coincide.
So: CI⊥DG