Mathematical geometry in the eighth grade of junior high school

(1) Proof: Because triangle ABC and triangle ADE are isosceles right triangles.

So AB=AC

Angle BAC=90 degrees

AD=AE

Angle DAE=90 degrees

Because angle BAD= angle BAC+ angle CAD = 90°+ angle CAD.

Angle CAE= Angle DAE+ Angle CAD = 90°+ Angle CAD

So bad angle = angle CAE

So triangle bad and triangle CAE congruence (SAS)

So BD=CE

(2) Proof: Because triangle BAD and triangle CAE are congruent (proved)

So angle ABM= angle ACM

Because angle ABM+ angle BAC+ angle ANB= 180 degrees.

So angle ACM+ angle ANB=90 degrees.

Because angle ANB= angle CMM

So angle ACM+ angle CNM=90 degrees.

Because angle ACM+ angle CNM+ angle CMN= 180 degrees.

So CMN angle =90 degrees.

So BD is perpendicular to CE.

(3) The conclusion is still valid.

The proof graph 1:: prolongs the intersection of DB and CE in F.

Because triangle ABC and triangle ADE are isosceles right triangles

So AB=BC

Angle EAC= angle BAD90 degrees

AD=AE

So triangle EAC and triangle BAD are congruent (SAS)

So BD=CE

Angle ACE= angle ABD

Because angle ABD= angle EBF

So angle ACE= angle EBF

Because angle EAC+ angle ACE+ angle FEB= 180 degrees.

So angle FEB+ angle EBF=90 degrees.

Because angle FEB+ angle EBF+ angle EFB= 180 degrees.

So EFB angle =89 degrees.

So BD is perpendicular to CE.