From Gogeometry

Using

• Additional points
• Inscribed angles in a circle
• congruent triangles
• Thales theorem
• Concyclic points

∠DBC = ∠DBA+∠ABC = π/2+∠ABC
∠ABG = ∠ABC+∠BCG = ∠ABC+ π/2
=> ∠DBC = ∠ABG
=> Δ DBC is congruent to Δ ABG (SAS)
=> ∠BDC = ∠BAG and DC=AG

Define K the intersection of AG and BC
=> ∠BDC = ∠BAG => ∠BDK = ∠BAK => AKBDE are concyclic
=> ∠AKD = ∠ABD =  π/2

In the ΔACG : O in the middle of AC and N in the middle of CG
Thales => ON//AG and ON=AG/2
In the ΔACD :O in the middle of AC and M in the middle of AD
Thales => OM//CD and OM=CD/4
Therefore OM=ON

∠AKD = ∠ABD =  π/2 and ON//AG and OM//CD
Therefore ∠MON =  π/2