For the following figure, can you deduce from the information $AT=CT$ and$DT=\frac{1}{2}DB$ that quadrilateral$ABCD$ is a parallelogram. If so, what theorem can you use?

Consider the figure.

If two lines are cut by a transversal line and the sum of two interior angles on the same side of the transversal is supplementary then those two lines are parallel.

The corresponding parts of congruent triangle are congruent.

Consider $AT=CT;DT=\frac{1}{2}DB$.

Consider the triangles $\Delta ATB,\Delta CTD$ and thus

Since $\angle ATB$ and $\angle CTD$ are vertically opposite angles thus they are equal such that $\angle ATB=\angle CTD$.

Use the SAS postulate as follows:

$\Delta ATB\cong \Delta CTD$

Since corresponding part of congruent triangle are congruent such that .

$\Delta ATB\cong \Delta CTD$

Thus $AS=BQ$.

Since “If two lines cut by a transversal line and sum of two interior angles on the same side of the transversal are supplementary then that two lines are parallel” thus$\overline{AB}$is parallel to $\overline{DC}$ such that$\overline{AB}\left|\right|\overline{DC}$ .

Since corresponding part of congruent triangle are congruent such that .

$\Delta BAT\cong \Delta DCT$

Thus, ABCD is a parallelogram.