In trapezoid ABCD, AB / / CD, de / / BC, CD = 7cm, the perimeter of trapezoid is 35cm, and that of triangle ade is 35cm

In trapezoid ABCD, AB / / CD, de / / BC, CD = 7cm, the perimeter of trapezoid is 35cm, and that of triangle ade is 35cm

CD is the top and bottom, e is on ab
Because: DC / / EB and de / / BC
So: quadrilateral DEBC is parallelogram
So: DC = EB = 7 CB = de
Because the trapezoid perimeter = AD + AE + EB + BC + DC = AD + AE + BC + 7 + 7 = 35
Because the perimeter of triangle ade = AD + AE + de = AD + AE + BC = 35-14 = 21

In the trapezoid ABCD, AB / / DC, CE are bisectors of angle BCD, and CE is perpendicular to ad, de = 2ae. CE divides the trapezoid into two parts, S1 and S2
In the trapezoid ABCD, AB / / DC, CE are bisectors of angle BCD, and CE is perpendicular to ad, de = 2ae. CE divides the trapezoid into two parts: S1 and S2. If S1 = 1, S2 is calculated

So OE = De, de = 2ae, so OA = AE. So OA: od = 1:4. AB / / DC, so the area of triangle OAB: the area of triangle OCD = 1:16, so the area of triangle OAB: ladder

Square trapezoid ABCD, ad ‖ BC, BD ⊥ DC, ∠ BCD bisector intersection BD and E, BD = 12, CE = 8 to calculate the area of trapezoid ABCD

Let BCD = 2x, ed = 8sinx in CDE, then be = 12-8sinx. In △ BEC, by the sine theorem, be / SiNx = CE / cos (2x), substituting be = 12-8sinx, CE = 8, then (12-8sinx) / SiNx = 8 / cos (2x), 4 (SiNx) ^ 3-6 (SiNx) ^ 2-4sinx + 3 = 0 is obtained

In the trapezoidal ABCD, AB is parallel to CD, and ∠ ADC + ∠ BCD = 90 degrees. Ad, AB and BC are the hypotenuse, and they are made into isosceles right triangles outward. Their areas are s1s2s3 and S1 + S3 = 4s2 respectively

The solution is BM ‖ ad through point B,
∵ ab ∥ CD, ∥ quadrilateral admb is a parallelogram,
∴AB=DM,AD=BM,
And ∵ ∠ ADC + ∠ BCD = 90 °,
That is to say, △ MBC is RT △,
∴MC2=MB2+BC2,
Take ad, AB and BC as the hypotenuse to make equal right triangle outwards,
∴△AED∽△ANB,△ANB∽△BFC,
$\frac{{S}_ {1}}{{S}_ {2}}$=$\frac{{AD}^{2}}{{AB}^{2}}$,$\frac{{S}_ {2}}{{S}_ {3}}$=$\frac{{AB}^{2}}{{BC}^{2}}$,
Ad2 = $- frac {s}_ {1}{AB}^{2}}{{S}_ {2}}$,BC2=$\frac{{S}_ {3}{AB}^{2}}{{S}_ {2}}$,
∴MC2=MB2+BC2=AD2+BC2=$\frac{{S}_ {1}{AB}^{2}}{{S}_ {2}}$+=$\frac{{S}_ {3}{AB}^{2}}{{S}_ {2}}$=$\frac{{AB}^{2}（{S}_ {1}+{S}_ {3}）\;}{{S}_ {2}}$,
∵S1+S3=4S2,
Ψ MC2 = 4ab2, MC = 2Ab, Gu xuanb