Extend EM and cross OB to n.

∫A coordinate (-4,0), B coordinate (0,4)

∴OA=OB=4

∫M is the midpoint of BF.

∴MB=MF

∵OB⊥OA EF⊥OA

∴EF∥OB

∴∠MBO=∠MFE ∠MNB=∠MEF

∴△MBN≌△MFE(AAS)

Me = mnbn = ef (i.e. BN=AE).

And OA⊥OB

∴OM=EM (the midline of the hypotenuse of a right triangle is equal to half of the hypotenuse)

OB-BN=OA-AE

∴OE=ON

∴ OM⊥EM (center line of the bottom of isosceles triangle, perpendicular to the bottom)

∴OM=EM,OM⊥EM

(2)OM=EM, OM⊥EM remains unchanged for the following reasons:

Extend EM to n, make MN=EM, connect BN, OE, o n,

And MB=MF (confirmed) ∠EMF=∠NMB (equiangular).

∴△MBN≌△MFE(SAS)

∴BN=EF (i.e. BN=AE)

∠∠BNE =∠ finite element method

∴BN∥EF

AE⊥EF again.

∴AE⊥BN

And OA⊥OB

∴∠EAO=∠NBO

(If two sides of one acute angle are perpendicular to two sides of the other acute angle, then the two acute angles are equal. )

And AE=BN OA=OB.

∴△EAO≌△NBO(SAS)

∴∠EOA=∠NOB

∠∠AOB =∠EOA+∠EOB = 90

∴∠NOB+∠EBO=90

That is EO ⊥ no EO = no.

∴△EON is an isosceles right triangle.

And MN=EM.

∴OM=EM OM⊥EM

(3) Extend OG to K, so that GK=GO, then OK=GK+GO=GA+GO.

It's ∵GA⊥OK again.

∴△AGK is an isosceles right triangle, that is, AK: GA = root number 2, ∠ PAO = 45.

In isosceles Rt△AOB: P is the midpoint of hypotenuse AB, then OP⊥AB and OP=AP.

∴△AOP is an isosceles right triangle

∴ ao: AP = root number 2 ∠ kag = 45.

∴AO:AP=AK:AG ∠PAO=∠KAG

∴∠PAO+∠GAO=∠KAG+∠GAO

That is ∠PAG =∠ Oak.

∴△PAO∽△OAK

∴ ok: gp = OA: AP = root number 2

That is, (GA+GO)/GP = root number 2.

I am glad to answer the above questions for you, and I hope it will be helpful to your study!