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Mathematical geometry problem
Solution: ① Make DY⊥AG at point D and cross AG at point Y.

CD = CE = DE = 2cm

△ CDE is an equilateral triangle.

∴∠CDE=∠DCE=∠DEC=60

∴∠adg=360-∠ADC-∠gde-∠CDE = 360-90-90-60 = 120

∴∠day=( 180- 120)× 1/2 = 30

∴dy= 1/2ad= 1/2× 1=0.5(cm)

That is, the distance from point d to AG is 0.5cm.

②? ∵α=45

∴∠NCE=∠NEC=45

∴∠CNE=90 ∴∠HND=∠CNE=90

∴∠HND=∠D=∠H=90

∴ Quadrilateral HNDM is a rectangle (a quadrilateral with three right angles is a rectangle)

CD = EH (known) CN=EN (confirmed)

∴CD-CN=EH-EN, that is, HN=DN.

∴ Quadrilateral HNDM is a rectangle (a group of rectangles with equal adjacent sides is a square)

Also, there is a small mistake in the above picture: