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Ad is the median line of △ ABC, EF is the median line of △ ABC, then the quantitative relationship between EF and ad is__________ .
E. F is the midpoint of AB and AC respectively
D is the midpoint of BC,
ED∥AC
EF∥BC
FD∥AB
Quadrilateral aefd is a parallelogram
Diagonal EF and ad, equally divided
In trapezoid ABCD, AD / / BC, EF are median lines, the area of trapezoid is equal to 48CM ^ 2, the area of triangle AEF is calculated
If the height of trapezoid is made through a, let BC intersect with G, and EF intersect with H
The area of trapezoid is (AD + BC) * Ag / 2 = 48
The area of AEF is EF * ah / 2
Because EF = (AD + BC) / 2
AH=AH/2
So the area of AEF is 1 / 4 of the trapezoid area
That's 12 cm ^ 2
In the isosceles trapezoid ABCD, the two diagonal lines AC and BD are perpendicular to each other, the median EF is 8cm long, and the height ch is calculated
Make AC parallel line through B and intersect DC extension line at o
Because AC is perpendicular to BD, Bo is perpendicular to BD
Co = AB, then do = 2ef = 16
bd=bo
BOD is an equilateral right triangle, the easy right side ob = BD = 8, radical 2
Triangle area = 64 = trapezoid area
Area divided by median = height = 8cm
That's it. Some simple processes are omitted in the middle. You can make up for them yourself
As shown in the figure, in the isosceles trapezoid ABCD, AB / / DC EF is the median line, and EF = 15 ∠ ABC = 60 BD bisection ∠ ABC is used to calculate the circumference of the trapezoid
∵∠ ABC = 60, BD bisects ∠ ABC
∴∠ABD＝∠CBD＝∠ABC/2＝30
∵AB//DC
∴∠CDB＝∠ABD＝30
∴∠CDB＝∠CBD
∴CD＝BC
∵ isosceles trapezoid ABCD
∴AD＝BC,∠A＝∠ABC＝60
∴∠ADC＝180-∠A-∠ABD＝90
∴AB＝2AD
∴AB＝2BC＝2CD
∵ trapezoidal median EF = 15
∴AB+CD＝2EF＝30
∴2CD+CD＝30
∴CD＝10
∴AD＝BC＝10,AB＝20
Ψ trapezoid perimeter = AB + BC + CD + ad = 20 + 10 + 10 + 10 = 50
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