E. F is the key points of AD and BC on the sides of ABCD, and G and H are the midpoint of BD and AC fast

E. F is the key points of AD and BC on the sides of ABCD, and G and H are the midpoint of BD and AC fast

GF, FH, he and eg are successively connected to form a quadrilateral gfhe, because he is the median line of ACD, he is parallel and equal to half of CD, GH is the median line of DBC, and FG is parallel and equal to half of CD
FG and he are parallel and equal. It can be proved that the quadrilateral gfhe is a parallelogram, and the diagonals of the parallelogram are equally divided with each other, so EF and GH are equally divided with each other

It is known that E.F. is a quadrilateral ABCD edge AD.bc The midpoint, G.H. is the midpoint of BDAC

By connecting eg, GF, FH and he, we can see that they are the median lines of DAB, BDC, cab and ADC. According to the median line theorem, we can see that GF is parallel and equal to half of DC, EH is parallel and equal to half of DC, namely GF ‖ eh; eg is parallel and equal to half of AB, HF is parallel and equal to half of AB, namely eg ‖ HF

It is known that Mn is the median line of ladder ABCD. AC and BD intersect with Mn respectively. F, e, ad = 30cm, BC = 40cm and the length of EF

Proof by similarity
Because Mn / / ad
So △ BMF ∽ bac
So ad ∶ MF = ab ∶ am = 2 ∶ 1
Similarly, △ CAD ∽ Cen
So ad ∶ endc ∶ DN = 2 ∶ 1
So MF = en = 15
Because Mn is the median line
So Mn = 15
So EF = (MF + EN) / 2 = (15 + 15) / 2 = 15
I don't know your graph. I put the intersection E on the left side of F and give more points!

In trapezoidal ABCD, ad is parallel to BC, AC is vertical to BD, EF is trapezoidal median line ∠ DBC = 30 ° and EF = AC

Set AC to BD at h,
BC = 2hc is obtained from AC vertical BD, ∠ DBC = 30 °;
From the parallel BC of AD, we can get ∠ BDA = 30 ° and ad = 2ah;
From EF to trapezoidal median line,
We get EF = (BC + AD) / 2
.=(2HC+2AH)/2
=HC+AH
=AC