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The standard answer to question 5.6.9 in the eighth grade math exercise 19. 1 published by People's Education Press.
5. As shown in the figure, the diagonal AC and BD of ◢◤ABCD intersect at point O, and points E, F, G and H are the midpoint of ao, Bo, Co and Do respectively. It is proved that the quadrilateral EFGH is a parallelogram.

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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