91Ó°ÊÓ

The given diagram is:

From the given diagram it can be noticed that:

$∠ABC=∠DAB=90°$and $ABâˆ\yen CD$.

As, $ABâˆ\yen CD$, therefore the angles $∠ABC$and $∠BCD$are the angles on the same side of the transversal BC.

Therefore, the sum of the angles $∠ABC$and $∠BCD$is $180°$.

Therefore, it can be obtained that:

$\begin{array}{c}∠ABC+∠BCD=180°\\ 90°+∠BCD=180°\\ ∠BCD=180°−90°\\ ∠BCD=90°\end{array}$

Therefore, the measure of the angle $∠BCD$is $90°$.

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

Therefore, it can be obtained that:

$\begin{array}{c}m∠ABC+m∠BCD+m∠CDA+m∠DAB=360°\\ 90°+90°+m∠CDA+90°=360°\\ m∠CDA+270°=360°\\ m∠CDA=360°−270°\\ m∠CDA=90°\end{array}$

Therefore, the measure of the angle $∠CDA$is $90°$.

Therefore, it can be noticed that $∠BCD=∠DAB=90°$and $∠ABC=∠ADC=90°$.

Therefore, $∠BCDâ‰\dots ∠DAB$.

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

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

Therefore, the given quadrilateral ABCDis 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.