It is proved that∵ quadrilateral ABCD is a parallelogram.
∴AO=CO,BO=DO
Points e, f, g and h are the midpoint of ao, bo, co and do respectively.
∴oe= 1/2oa,of= 1/2ob,og= 1/2oc,oh= 1/2od
∴OE=OG,OF=OH
∴ quadrilateral EFGH is a parallelogram.
6. As shown in the figure, quadrilateral ABCD is a parallelogram, ∠ABC = 70, BE bisector ∠ ABC and AD intersect at point E, and DF∨BE and BC intersect at point F .. Find the size of < 1.
Solution: ∫ quadrilateral ABCD is a parallelogram, ∠ ABC = 70.
∴AD∥BC,ED∥BF
And ∵df∨be ∴ebfd is a parallelogram, ∠EBF=∠EDF.
∫BE divided equally ∠ABC, ∠ ABC = ∠ ADC = 70.
∴∠ EBF =1/2 ∠ ABC =1/2x70 = 35 degrees, and French power = 35 degrees.
∴∠ 1=∠ADC-∠EDF=70 -35 =35
9. In quadrilateral ABCD, AD= 12, DO=OB=5, AC=26, ∠ ADB = 90. Find the length of BC and the area of quadrilateral ABCD.
Solution: At △AOD, ∠ ADB = 90, AD= 12, 0D=5,
According to Pythagorean theorem, draw a conclusion.
OA2 = OD2+AD2 = 52+ 122 = 169，
∴OA= 13.
AC = 26，OA= 13，
∴OA=OC.
DO=OB again,
∴ quadrilateral ABCD is a parallelogram.
∴AD=BC= 12.
∫∠ADB = 90，
∴AD⊥BD.
∴S quadrilateral ABCD=AD? BD= 12× 10= 120。
Answer: The length of BC is 12, and the area of quadrilateral ABCD is 120.
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