14. Given the <DAF =100° What must be the m<BCD to prove that quadrilateral ABCD is a parallelogram?
A. 20°
B. 80°
C. 90°
D. 100°
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14. Given the <DAF =100° What must be the m<BCD to prove that quadrilateral ABCD is a parallelogram?
A. 20°
B. 80°
C. 90°
D. 100°
help po pls
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[tex]\huge\red{\overline{\qquad\ qquad\qquad\qquad\qquad}}[/tex]
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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