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Math and geometry proving problem in the first grade of junior high school.
Solution: ∫ bisected ∠ABC (known)

∴∠ABC=2∠ 1 (definition of angular bisector)

∫CE bisection ∠DCB (known)

∴∠DCB=2∠2 (definition of angular bisector)

∴∠abc+∠dcb=2∠ 1+2∠2=2(∠ 1+∠2)

∫≈ 1+∠2 = 90 (known)

∴∠ABC+∠DCB=2×90 = 180,

∴AB∥CD (complementary to the inner corner on the same side, two straight lines are parallel).