91Ó°ÊÓ

The given diagram is:

From the given diagram it can be noticed that:

$∠ABC=∠ADC=70°$.

Therefore, $∠ABCâ‰\dots ∠ADC$.

As we know that the sum of the angles of a quadrilateral is $360°$.

Therefore, it can be obtained that:

$\begin{array}{c}m∠ABC+m∠BCD+m∠CDA+m∠DAB=360°\\ 70°+m∠BCD+70°+110°=360°\\ m∠BCD+250°=360°\\ m∠BCD=360°−250°\\ m∠BCD=110°\end{array}$

Therefore, the measure of the angle $∠BCD$is $110°$.

Therefore, it can be noticed that $∠BCD=∠DAB=110°$.

Therefore, $∠BCDâ‰\dots ∠DAB$.

Therefore, it can be said that $∠ABCâ‰\dots ∠ADC$and$∠BCDâ‰\dots ∠DAB$

That implies that both pairs of opposite angles of the quadrilateral are congruent.

Therefore, the given quadrilateral ABCD is a parallelogram.

Yes, the given figure ABCD is a parallelogram.

The theorem that applies is that if both pairs of opposite angles of the quadrilateral are congruent then the quadrilateral is a parallelogram.