Math problems in the senior high school entrance examination! ! ! ! ! ! !

(1) According to the meaning of the question, the quadrangle ECDO is a rectangle.

So CD is parallel, equal to OE, so angle CEO= angle CDE.

Because OG=EH

So the triangle OEH is equal to the triangle CDG(SAS), so OH=CG.

Similarly, the triangle CEH is equal to the triangle ODG, so HC=OG.

So the quadrilateral OGCH is a parallelogram.

(2) The length of 2)DG remains unchanged.

Connecting company

Because the quadrilateral CDOE is rectangular, CO=ED= radius.

So GD= 1/3OA= 1.

(3) 12

G crossing makes GM perpendicular to AO and crosses m.

Original formula =ED2-CE2+3(OM2+MG2)

=ED2-CE2+3(OM2+GD2-MD2)

=9-CE2+3(OM2+ 1-MD2)

=9-CE2+3+3(MO+MD)(MO-MD)

= 9-CE2+3+3d (Mo-Mo)

= 12+CE(3MO-3MD-3MO-3MD)

12+CE(2MO-4MD)

Because the triangle MGO is similar to the triangle DEO

So OM=2MD

So 2MO-4MD=0.

So the original formula = 12.