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14. Given the <DAF =100° What must be the m<BCD to prove that quadrilateral ABCD is a parallelogram?

D. 100°

[tex]\sf \green{Example}[/tex]

Given ABCD is a parallelogram.

The sum of co-interior angles is always 180°.

Given ∠B=100°

and, ∠B+∠C=180°

⇒∠C=80°

The opposite angles of a parallelogram are equal

i.e, ∠A=∠C=80°

∴ Sum ∠A+∠C=80°+80°=160°

Hence, the answer is 160°

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