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Math and geometry problems in the second day of junior high school?
(1) proves that AD is perpendicular to BC and D.

Therefore, angle ADB= angle BDF= angle ADC=90 degrees.

Because angle ABC+ angle ADB+ angle BAD= 180 degrees.

Angle ABC=45 degrees

So the angle BAD=45 degrees

So angle BAC= angle BAD=45 degrees.

So AD=BD

Because BE is perpendicular to AC and e.

So BEC angle =90 degrees.

Because angle BEC+ angle ACD+ angle CBE= 180 degrees.

So angle ACD+ angle CBE=90 degrees.

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

So angle CBE+ angle BFD=90 degrees.

So angle BFD= angle ACD

Angle BDF= angle ADC=90 degrees

So triangular BDF and triangular ADC are congruent (AAS)

So DF=DC

So the triangle CDF is an isosceles triangle

Because angle ADC= angle CDF=90 degrees

So the angle CDF=90 degrees.

So the triangle CDF is an isosceles right triangle.

(2) Proof: Because AD is perpendicular to BC and D ..

So angle ADB= angle BDF= angle CDA=90 degrees.

Because angle ABC+ angle ADB+ angle BAD= 180 degrees.

Angle ABC=45 degrees

So the angle BAD=45 degrees

So angle BAD= angle ABC=45 degrees.

So AD=BD

Because angle ADC+ angle CAD+ angle ACB= 180 degrees.

So angle ACB+ angle CAD=90 degrees.

BeCAuse BE is perpendicular to AC, the extension line of ca is in e.

So BEC angle =90 degrees.

Because BEC angle +FBD angle +ACB angle = 180 degrees.

So FBD angle +ACB angle =90 degrees.

So angle FBD= angle CAD

Angle BDF= angle ADC=90 degrees

So triangular BDF and triangular ADC are congruent (AAS)

So DF=CD

Because angle ADC= angle CDF

So the angle CDF=90 degrees.

So the triangle CDF is an isosceles right triangle.