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Find a math problem of similar triangles in the second day of junior high school.
Proof: Because △ABC and △DEC are isosceles right triangles.

So CD/CE = AC/CB =1√ 2.

DCE=∠ACB=45 degrees

So ∠DCA=∠ECB.

According to Nordic Airlines

△ACD≈△BCE

2) Because △ACD≈△BCE

So ∠DAC=∠EBC=45 degrees.

So ∠ DAB+∠ CBA = ∠ DAC+∠ cab+∠ CBA = 45+90+45 =180 degrees.

So in ∨ BC,