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Seeking process. Mathematics rotation problem in grade three
25. (1) Proof: Turn the triangle BEC counterclockwise by 90 degrees to get the triangle BAF and connect it with DF.

So triangle BEC and triangle BAFF are congruent.

So FA=EC

BF=BE

Angle ABF= angle CBE

Because the angle ABC=90 degrees

AB=AC

So the triangle ABC is an isosceles right triangle.

So angle BAD= angle BCE=45 degrees

So the angle BAF=45 degrees

Because angle ABC= angle ABD+ angle DBE+ angle CBE=90 degrees.

Angle DBE=45 degrees

So angle ABF+ angle ABD= angle DBF=45 degrees.

So angle DBF= angle DBE=45 degrees.

Because BD=BD

So triangle DBF and triangle DBE are congruent (SAS)

So DF=DE

Because angle DAF= angle BAF+ angle BAD=45+45=90 degrees.

So the triangle DAF is a right triangle.

From Pythagorean Theorem:

DF^2=AF^2+AD^2

So de 2 = ad 2+EC 2.

(3)MN=AM+CN

Proof: extend DA to make AF=CN and connect BF.

Because the quadrilateral ABCD is a square

So angle BAF= angle BCN=90 degrees.

Angle ABC= Angle ABM+ Angle MBN+ Angle CBN=90 degrees.

AB=BC

So triangular Baff and triangular BCN are congruent (SAS)

So BF=BN

Angle ABF= angle CBN

So angle ABF+ angle ABM+ angle MBN=90 degrees.

Because the angle MBN=45 degrees

So ABF angle ABM angle = MBF angle =45 degrees.

So angle MBF= angle MBN=45 degrees.

Because BM=BM

So triangle BMF and triangle BMN are congruent (SAS)

So MF=MN

Because MF=AM+AF

So MN=AM+CN