20. In quadrilateral $\mathrm{ABCD}$, $\angle B=90^{\circ}$, and $A B=B C=5, C D=2 \sqrt{15}, D A=\sqrt{10}$, the degree measure of $\angle B A D$ is
(A) $90^{\circ}$
(B) $120^{\circ}$
(C) $135^{\circ}$
(D) $150^{\circ}$
Answer: (C)
Solution: As shown in the figure below, connect $\mathrm{AC}$,
Since
$$
\angle B=90^{\circ}, A B=B C=5 \text {, }
$$

Therefore
$$
A C=\sqrt{A B^{2}+B C^{2}}=5 \sqrt{2}, \angle B A C=45^{\circ} \text {, }
$$

Also
$$
C D=2 \sqrt{15}, D A=\sqrt{10},
$$

Therefore
$$
A C^{2}+D A^{2}=50+10=60, C D^{2}=60,
$$

Thus
$$
A C^{2}+D A^{2}=C D^{2},
$$

Hence $\triangle \mathrm{ACD}$ is a right triangle, and $\angle C A D=90^{\circ}$,

Therefore
$$
\angle B A D=45^{\circ}+90^{\circ}=135^{\circ} \text {, }
$$