Question

in quadrilateral ABCD, AB parallel CD and AD = BC. prove that angel A = angel B.

Answer 1

Drop an altitude onto AB from D and from C. Since AB∥DC, the altitudes have the same length. Also AD=BC, so

sin∠A=sin∠B

Either ∠A=∠B, or ∠A+∠B=180∘.

Let E∈DC such that AE||BC. Thus, AECB is parallelogram,

which gives AE=BC and from here AE=AD.

Thus, ∡ADE=∡AED=∡BCD.

Now,

∡DAB=180∘−∡ADC=180∘−∡BCD=∡ABC

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