Through E and B, BQ is parallel to CE(C and H coincide), EQ is parallel to BC, QC is connected, and GE extends to M.

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