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Mathematical geometry problem
The condition is poor. If an AC vertical BD is added, the conclusion can be solved.

Method 1: Half of diagonal product

Solution: Because the quadrilateral ABCD is an isosceles trapezoid.

So AC=BD

Parallel BC

Because DE is in parallel with AC

So the quadrilateral is a parallelogram.

So AC=DE

AD=CE=3

BD vertical DE

So the angle BDE=90 degrees

So the triangle BDE is a right triangle.

From Pythagorean Theorem:

BE^2=BD^2+DE^2

Because BE=BC+CE

BC=7

BD=DE

So BD=DE=5 times the root number 2.

So S- trapezoidal ABCD= 1/2*AC*BD=25.

Method 2: Cutting repair method

Solution: Because the quadrilateral ABCD is an isosceles trapezoid.

So AC=BD

Parallel BC

Because DE is in parallel with AC

So AD=CE

AC=DE

S triangle ACD=S triangle DCE=S triangle ABD

BD vertical DE

So the angle BDE=90 degrees

From Pythagorean Theorem:

BE^2=BD^2+DE^2

Because BE=BC+CE

BC=7

AD=3

So BD=DE=5 times the root number 2.

Because s trapezoid ABCD=S triangle ABD+S triangle BDC.

S triangle BDE=S triangle BDC+S triangle DCE

So s trapezoid ABCD=S triangle BDE

Because the S triangle BDE= 1/2*BD*DE=25.

So the S-trapezoid ABCD=25.

Method 3: (upper bottom+lower bottom) times the height and divides by 2.

Solution: the intersection d is DF, perpendicular to BC and F.

So BFD angle =90 degrees.

From Pythagorean Theorem:

BD^2=DF^2+BF^2

Because the quadrilateral ABCD is an isosceles trapezoid

So AC=BD

Parallel BC

Because DE is in parallel with AC

So the quadrilateral is a parallelogram.

So AD=CE

AC=DE

Because AC vertical BD

So BD is perpendicular to DE.

So the angle BDE=90 degrees

BD=DE

So the triangle BDE is an isosceles triangle.

So DBF angle =45 degrees.

Derived from Pythagorean theorem;

BE^2=BD^2+DE^2

Because BE=BC+CE

AD=3

BC=7

So BD=DE=5 times the root number 2.

Because DBF angle +BFD angle +BDF angle = 180 degrees.

So DBF angle = BDF angle =45 degrees.

So DF=BF

So DF=5

Because S- trapezoid ABCD= 1/2*(AD+BC)*DF=25.

So the S-trapezoid ABCD=25.