Mathematics for junior 2 quadrilateral ABCD, AB BC 90 degrees, BE perpendicular AD, and the area of

Solution: Make an extension of the BF CD to F

be ad again; d=90°, then the quadrilateral bfde is rectangular.

Therefore ebf=90°= abc, cbf= abe;

ab=bc; BCF = BEA = 90 degrees. then BCF δbea(AAS)

So: bf=be.That is, the quadrilateral bfde is a square.

s bcf = s bea, s square bfde=s quadrilateral abcd

So be= 8=2 2

Whether the cd is perpendicular to the AD, otherwise it cannot be calculated.

Extending DC to point F to connect BF so that BF AD is then proven to be AEB congruent with BFC.

The angle abc is equal to 90 degrees, right?

Does the original question say that the angle d is equal to 90 degrees?

Do CF BE, pass BE to F, AED= BFC, AB=BC, liquid denier AED antipulsation BFC(HL Rule).

be=cf,ae=bf.Bridge disturbance = def== efc== efc, quadrilateral defc is rectangular, cf=de.

s△abcd=s△aeb+s△bfc+de*ef=be*ae+de*ef=be*ae+de*(be-bf)=be*ae+de*(be-ae)=de*be=be^2=16,∴be=4

As CG perpendicular to point G, easy to prove Abe BCG

Rotate ABE around point B by 90° to ACF to get a square EBFD, then the area of the square = the area of the original quadrilateral = 8

So be=8=2 2

From these ab = bc, angle abc = angle cda = 90 degrees can be obtained, the quadrilateral abcd is the square, then point e is point a, that is to say, the two points coincide, the square area formula is the side length multiplied by the side length, so be is 2 times the root number 2

Passing point B to make a BF extension line perpendicular to DC at point F proves that ABE CBF

AAS), the area of the square EBFD is equal to the area of the quadrilateral ABCD that is equal to 8, so BE= 8=2 2

1) Move abc to bfc place ab=bc

The BC side is in the BF position.

bf=be quadrilateral, bfed is square.

The area of the quadrilateral ABCD is 8

s bfed=s quadrilateral, abcd=8

be=bf=root number 8

be=bf=

The drawings are not good, I can't understand them.

Let's make a simple :::, as CG vertical be to point G, easy to prove Abe BCG

Rotate ABE around point B by 90° to ACF to get a square EBFD, then the area of the square = the area of the original quadrilateral = 8

So be=8=2 2

Solution: pass point B as the perpendicular line on the DC extension line, so that the perpendicular foot is F, angle AEB = angle BFC = 90 degrees; According to the inner angle and 360° of the quadrilateral, it can be concluded that the angle BAE = the angle BCF; BC=AB, so the triangle BFC is congruent with BEA. So: bf=be.

The area of the quadrilateral is converted into the area of the square bedf. So be = root number 8 = 2 times root number 2

Straight line that crosses point B and runs in AD, from a to make a perpendicular line to f, extend DC to G, then bedg is a rectangle, because the angle abc = 90 degrees, the angle ebg = 90 degrees, so the angle abe = angle cbg

Again, ab=bc, right triangle, bgc abe=bg, so the quadrilateral ebgd is square and the area of the square ebgd = 8

Wax contains be=2 2

Solution: Make CG perpendicular to point G, easy to prove Abe BCG

Rotate ABE around point B 90° to the lead deficiency ACF to obtain a square EBFD, then the area of the square or = the area of the original quadrilateral = 8

So be=8=2 2

There are no points to offer, so I won't write in detail.

Passing the point C to make the be perpendicular line, the perpendicular foot is F

It is easy to obtain the congruence of triangle BAE and CBF (from the relationship between ab=ac and co-angle, you should be able to prove it).

So there is bf=ae be=cf

So the total area is:

ae*be 2+bf*cf 2+de*ef=8 unify equal lines.

be*bf+be*（be-bf）=8

So be = 2 times the root number 2