The first question:
According to "E and H are the midpoint of AB and AD respectively, F and G are points on BC and CD respectively, CF/CB = CG/CD = 2/3, BD = 6 cm", the lengths of EH and FG are calculated (this is the knowledge of junior high school), and then the height of trapezoid, that is, the distance between EH and FG, is calculated by the area formula of trapezoid.
The second question:
This question needs to be used as an auxiliary line. Take the midpoint G in AC, connect EG and FG, and find the angle formed by EF AD=BC, which is the angle FEG. Then, according to the condition that "E and F are the midpoint of AB and CD respectively, AD is perpendicular to BC, and AD = BC", it can be found that the triangle EFG is an isosceles right triangle.
Good understanding, pay attention to draw a sketch, believe me.
(1) solution:
Make EM perpendicular to FG and point m, so the distance between EH and FG is EM.
In triangle ABD, because E and H are the midpoint of AB and AD,
So EH is parallel, equal to 1/2BD, which means EH=3.
Because trapezoid area =(EH+FG)* EM* 1/2.
Get (3+4)*EM* 1/2=28。
That is, EM=8.
Therefore, the distance between parallel lines EH and FG is 8.
(2) Solution:
Set the AC midpoint to n and connect EN and NF.
In triangle ABC, EG is the center line, so EG= 1/2BC.
In a delta ADC, NF is the zero line, so NF= 1/2AD.
According to the meaning of the question, AD is vertical and equal to BC.
The obtained NF is perpendicular to en and NF=EN.
Therefore, EFN is an isosceles right triangle,
That is, the angle between EF and BC is 45.
Whoo-hoo! I'm exhausted. You should take a closer look.
Cut out the main information and draw a picture. Take your time ~ ~ Mine should be very detailed, right?
Hmm. How interesting