In the isosceles trapezoid ABCD, ad ∥ BC, AB is equal to DC, AC and BD intersect at the point O, ∠ BOC is equal to 90 ° and BD = 10?

In the isosceles trapezoid ABCD, ad ∥ BC, AB is equal to DC, AC and BD intersect at the point O, ∠ BOC is equal to 90 ° and BD = 10?

Isosceles trapezoid can deduce AC = BD = 10, the diagonal is vertical, we regard this trapezoid as two triangles ACD ACB. The total area is equal to the sum of the two triangles area = 1 / 2Ac (OD + OB) = 1 / 2Ac * BD = 50

In trapezoid ABCD, AD / / BC AB = DC ad = 2 BC = 4 AC vertical BD angle BOC = 60 degrees to calculate trapezoid area

Is o the intersection of AC and BD
If it is AC ⊥ BD
h=4/2+2/2=3
S=(2+4)×3÷2=0
If ∠ BOC = 60 degree
h=3√3
S=9√3

In the parallelogram ABCD, the diagonal AC and BC intersect o, BD = 2ad, e, F and G are the midpoint of OC, OD and ab respectively, and eg = EF

It is proved that: ∵ E and F are the midpoint of OC and OD respectively, ∵ EF is the median line of ⊿ OCD, ∵ EF = (# 189); CD connects be, ∵ ABCD is a parallelogram ∵ ad = BC, Ao = OD, ab = CD ∵ BD = 2ad ∵ BC = Bo, that is, ⊿ CBO is an isosceles triangle ∵ be is the middle line of ⊿ CBO (the middle line of isosceles triangle, the high line and the angle dividing line are united) ≁ be ⊥ a

Parallelogram ABCD diagonal AC.BD If AB = a, BC = B, be = C, find the length of BF

Go through O and make og / / BC intersection AB with G, because o is the intersection point of parallelogram diagonal. According to the theorem of triangle median line, og = 1 / 2 * ad = 1 / 2 * BC = 1 / 2 * B,
BG=1/2*AB=1/2*a
The triangle EBF is similar to the triangle ego
EB：EG=BF：OG
After substituting the above values, we get:
c:(C+1/2*a)=BF:(1/2*b)
BF=1/2*b*C/(C+1/2*a)
=bc/(2C+a)