Parallelogram ABCD, AC = 2gc, AE = EF = FB, triangle GEF area is 6 square centimeter, find the area of parallelogram

Parallelogram ABCD, AC = 2gc, AE = EF = FB, triangle GEF area is 6 square centimeter, find the area of parallelogram

As shown in the figure, let EF = a cm and the height of △ EFG be h cm, then the area of △ EFG is: (1 / 2) ah & nbsp; = & nbsp; 6, then: ah & nbsp; = & nbsp; 12, the side of parallelogram: ab = 3, EF = 3A, and the height of parallelogram is 2 h. therefore, the area of parallelogram is: 3A × 2 h & nbsp; = & nbsp; 6 ah & nbsp; = & nbsp; 72 square cm

As shown in the figure, in the parallelogram ABCD, AC and BD intersect at point O, where e, F, G and H are the midpoint of AB, ob, CD and OD respectively

The diagonals AC and BD of parallelogram ABCD intersect at point O. points E and F are the midpoint of OB and OD respectively. Any straight line passing through point O intersects at AB and CD intersects at points g and h respectively
Verification: GF parallel EH

Connecting Ge, FH, because AB / / CD, OB = od in parallelogram ABCD, so angle ABO = angle CDO, because angle gob = angle hod, OB = OD, so triangle bog is equal to triangle DOH, so og = Oh, because points E and F are the midpoint of ob, OD, ob = OD, so OE = of, because og = Oh, so gehf is parallelogram, so GF

It is known that in the parallelogram ABCD, the diagonal lines AC and BD intersect at the point O, BD = 2ad, e, F and G are the midpoint of OC, OD and ab respectively

It is proved that: (1) the ∵ quadrilateral ABCD is a parallelogram, ∵ ad = BC, BD = 2BO. From the known BD = 2ad, ∵ Bo = BC, and E is the midpoint of OC, ∵ be ⊥ AC. (2) from (1) be ⊥ AC, and G is the midpoint of AB, ∵ eg is the midline on the hypotenuse of RT △ Abe. ∵ eg = 12ab. And ∵ EF is the median of △ OCD, ∵ EF = 1