Given that ad is parallel to BC and E is the midpoint of Cd in trapezoidal ABCD, is the area of trapezoidal ABCD twice that of triangular Abe

Given that ad is parallel to BC and E is the midpoint of Cd in trapezoidal ABCD, is the area of trapezoidal ABCD twice that of triangular Abe

It's twice that
Suppose ABCD is a right angled trapezoid, ad is perpendicular to AB, BC is perpendicular to AB, passing through e, EF is perpendicular to AB, and the intersection is f, then the area of triangle Abe is: EF * AB * 1 / 2
The area of trapezoidal ABCD is: (AD + BC) * AB * 1 / 2 = 2ef * AB * 1 / 2 = EF * ab
So the area of the trapezoid is twice that of the triangle

As shown in the figure, in the right angle trapezoid ABCD, ∠ C = 90 °, ad ‖ BC, AD + BC = AB, e is the midpoint of CD. If ad = 2, BC = 8, calculate the area of △ Abe

In RT △ ABG, we know from Pythagorean theorem that ag2 = ab2-bg2 = (AD + BC) 2 - (BC-AD) 2 = 102-62 = 82, ｜ Ag = 8, so ah = GH = 4, ｜ s △ Abe = s △ AEF + s △ bef = 12ef · ah + 12ef · GH = 12ef · (ah + GH) = 12 EF•AG=12×5×8=20．

It is known that in trapezoidal ABCD, ab ‖ CD and E are the key points of BC. Let the area of triangle DEA be S1 and the area of trapezoidal ABCD be S2. The quantitative relationship between S1 and S2 is explored and the reasons are explained
The diagram is as follows: trapezoid, e is the midpoint of CB, connecting de and AE

The conclusion is: S2 = 2S1
It is proved that when De is extended, the extension line of AB intersects at point F
∵AB‖CD
∴∠CDE=∠F,∠C=∠EBF
∵BE=CE
∴△DCE≌△FBE
Ψ de = EF, s △ ADF = s ladder ABCD
∵DE=BE
∴S△ADE=1/2S△ADF
∴S1=1/2S△ADF=1/2S2
That is, S2 = 2S1