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Rectangular problems in mathematical geometry in the second volume of the eighth grade
1, ABCD is rhombic, so AC is vertical and divided equally to BD, so OD=4, OC=3, so CD=5.

Triangle COE is similar to triangle o CD because OE is perpendicular to CD and shares internal angle OCD.

OD: CD = 4: 5 OC: CD = 3: 5,OC=3。

So CE=9\5

2.AB= 1, BC=2, so BD= root 5, so BO= (root 5)\2.

Because BE is perpendicular to AC and angle ACB= angle DBC.

So the triangle BEC is similar to the triangle DBC.

DB: CD = BC: BE = (root 5): 1 and BC=2.

So BE=2 root 5\5

Because BO= root 5\2

So OE=9\20