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(1) quadrilateral ACE'D' can form a parallelogram with a=45 degrees.

Proof: Because the angle ACB=90 degrees.

So the triangle DCE is a right triangle.

DC=EC= 1

So the triangle DCE is an isosceles right triangle.

So de 2 = DC 2+EC2.

Angle CDE= Angle CED=45 degrees.

So DE= root number 2

Because AC=BC= root number 2

So AC=DE= root number 2

Because the triangle DCE rotates clockwise around point C by 1 degree, the triangle D'CE' is obtained.

So D'E'=DE= root number 2.

ACB angle =90 degrees.

Angle CE'D'= angle CED=45 degrees.

So AC=D'E''= root number 2.

Because angle α = α+angle β

A=45 degrees

So the angle ACE'= 135 degrees.

So angle ACE+ angle CE'D'= 180 degrees.

So AC is parallel to D''E'

So the quadrilateral is a parallelogram.

(2) The quadrilateral FGHI is a square

Proof: connect BD'AE', AE' crosses FI in Q, AE' crosses BD in P, and BD' crosses FG in M.

Because angle α = angle ACB+ angle β

Angle ACB=90 degrees

So angle ACE'=90 degrees+angle BCE'

Because angle BCD'= angle BCE'+ angle D'CE'

Angle D'CE'=90 degrees

So angle BCD'=90 degrees+angle BCE'

So angle α = angle β.

Because AB=BC

D'C=E'C

So triangle ACE' and triangle BCD' are congruent (SAS)

So AE'=BD'

Corner CAE = corner CBD

So a, b, p and c are four-point circles.

So CBP angle APB angle =90 degrees.

Because F, G, H and I are the midpoint of AB, BE', E' D' and AD' respectively.

Therefore, FG, GH and FI fi are the center lines of triangle ABE, triangle BE'E' and triangle ABD'

So FG= 1/2AE'

FG parallel AE'

So angle FGD'+ angle APB= 180 degrees.

So the angle FGD'=90 degrees

The same can be proved: GH= 1/2BD'

GH parallel BD'

FI= 1/2BD '

FI parallel BD'

So FG=FI

Angle GFI+ Angle FGD'= 180 degrees.

So GFI angle =90 degrees.

FI=GH

FI parallel BD'

So the quadrilateral FGHI is a parallelogram.

Because GFI angle =90 degrees (proved)

So the quadrilateral FGHI is a rectangle.

FG=FI (authentication)

So the quadrilateral FGHI is a square.

(3) Solution: Because angle ACB+ angle BCE'+ angle D'CE'+ angle ACD'=360 degrees.

Angle ACB=90 degrees

Angle D'CE'=90 degrees (confirmed)

A= 135 degrees

So the angle BCE'=45 degrees.

Because the angle CE'D'=45 degrees

So angle BCE'= angle CE'D'=45 degrees.

So BC and D'E are parallel.

Because BC=D'E'= root number 2.

So quadrilateral BCD'E' is a parallelogram.

So E'B=D'C= 1.

Parallel direct current

So the angle can be'+angle BCD'= 180 degrees.

Because angle BCD'= angle BCE'+ angle D'CE'=45+90= 135 degrees.

So the angle CBE'=45 degrees.

Because the angle ACB=90 degrees

AC=BC= root number 2

So the triangle ABC is an isosceles right triangle.

So the angle ABC=45 degrees.

AB^2=AC^2+BC^2

So AB=2

I because angle ABE'= angle ABC+ angle CBE'=45+45=90 degrees.

So the triangle ABE' is a right triangle.

So AE' 2 = AB 2+E' B 2

So AE= root number 5

Because FG= 1/2AE' (authentication)

So FG= root number 5/2.

Because the quadrilateral FGHI is a square.

So the square of s fghi = fg 2 = 5/4.

So the area of quadrilateral FGHI is 5/4.

(4) From Figure (2), it can be concluded that regardless of a (0