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In the quadrilateral ABCD, AD / / BC, M is the midpoint of AD, MB = MC? Figure: a -------- D / / M \ \ / / \ \ // \ \ B----------------C
Because: MB = MC
So: angle MBC = angle MCB
Because: ad is parallel to BC
So: angle AMB = angle MBC, angle DMC = angle MCB
So: angle AMB = angle DMC
Because: m is the midpoint of AD
So: am = DM
Because: MB = MC
So: according to the edge theorem, triangle AMB and triangle DMC are congruent
So: angle ABM = angle DCM
So: angle ABC = angle DCB
Because: ad is parallel to BC
So: ab = DC
So: quadrilateral ABCD is isosceles trapezoid
I can't type mathematical symbols. They are all written in Chinese characters. Let's have a look
In the quadrilateral ABCD, CD is parallel to AB, M is the midpoint of CD, and Ma = MB. It is proved that the quadrilateral ABCD is isosceles trapezoid
In △ mAb, from Ma = MB, we can know: angle BAM = angle MBA
If CD is parallel to AB, then BAM = AMC, MBA = BMD
And Ma = MB, MC = MD (M is the midpoint of CD)
So △ AMC is equal to △ BMD
So AC = BD
CD parallel ab
So the quadrilateral ABCD is isosceles trapezoid
The area of trapezoidal ABCD is 4, and M is the midpoint of CD. If am and BM are connected, the area of triangular ABM is 4
The answer is 2. If the building owner can extend the AC extension line of am to e, it is easy to prove that △ AMD ≌ EMC, then am = em
The same base is the same height as that of △ BEM, s △ ABM = 4 △ 2 = 2
The area of trapezoid ABCD is 20 square centimeters, M is the midpoint of CD, and the area of triangle ABM is calculated
Extend the extension line of am-bc to n,
Then △ ADM is equal to △ NCM
The area of AM = Mn, △ ABN = trapezoidal ABCD = 20 square centimeters
The area of Δ ABM = the area of trapezoidal ABCD / 2 = 10
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