91Ó°ÊÓ

Line segment$\stackrel{¯}{DA}$ is transversal to the parallel lines$\stackrel{¯}{AB}$ and$\stackrel{¯}{CD}$

$∠DAB$and$∠CDA$ forms a pair of same-side interior angles

Theorem 3-3: If two parallel lines are cut by a transversal then the same-side interior angles are supplementary.

$$\begin{gathered} m∠DAB+m∠CDA=180 \\ m∠DAB+116=180 \\ m∠DAB=180−116 \\ m∠DAB=64 \end{gathered}$$

Since, $\stackrel{→}{AK}$bisects $∠DAB$. So $∠KAB$is half of angle $∠DAB$

$$\begin{gathered} ∠KAB=\frac{∠DAB}{2} \\ ∠KAB=\frac{64}{2} \\ ∠KAB=32 \end{gathered}$$

Line segment$\stackrel{→}{AK}$ is transversal to the parallel lines$\stackrel{¯}{AB}$ and$\stackrel{¯}{CD}$

$∠KAB$and$∠DKA$ forms a pair of alternate interior angles

Theorem 3-2: If two parallel lines are cut by a transversal then the alternate interior angles are congruent.

$$\begin{gathered} ∠DKAâ‰\dots ∠KAB \\ m∠DKA=m∠KAB \\ m∠DKA=32 \end{gathered}$$

Thus,$m∠DAB=64,m∠KAB=32\text{and}m∠DKA=32$

Line segments$\stackrel{¯}{DA}\text{and}\stackrel{¯}{BC}$ are transversal to the parallel lines$\stackrel{¯}{AB}$ and$\stackrel{¯}{CD}$

Angles which can be find are$∠DAB,∠KAB,∠KAD\text{,and}∠DKA$

Angles which cannot be find are $∠AKB,∠KBA,∠KBC\text{,}∠BKC,\text{and}∠KCB$

Thus, to find $∠C$more information is needed.

Hence, we cannot conclude that$∠D$ and$∠C$ are supplementary.