13. Given a convex quadrilateral $ABCD$ whose four interior angles form an arithmetic sequence. In $\triangle ABD$ and $\triangle BCD$, if $\angle ABD = \angle CDB$, $\angle DAB = \angle CBD$, and the three interior angles of these two triangles also form an arithmetic sequence, then the maximum possible value of the sum of the two largest interior angles of quadrilateral $ABCD$ is ( ).
(A) 210
(B) 220
(C) 230
(D) 240
(E) 250
13. D.

Assume the four interior angles of the quadrilateral are
$$
\alpha, \alpha+d, \alpha+2d, \alpha+3d \text{.}
$$

Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have
$$
\alpha+\alpha+3d=\alpha+d+\alpha+2d=180^{\circ} \text{.}
$$

Given that $\angle ABD = \angle CDB$ and $\angle DAB = \angle CBD$, we know that $\triangle ABD \sim \triangle BDC$.
Let $\angle ABD = \angle CDB = \alpha_1$,
$\angle DAB = \angle CBD = \alpha_2$,
$\angle ADB = \angle BCD = \alpha_3$.
Then the four interior angles of the quadrilateral are $\alpha_2, \alpha_1 + \alpha_2, \alpha_3, \alpha_1 + \alpha_3$.
Since the three interior angles of a triangle form an arithmetic sequence, one of $\alpha_1, \alpha_2, \alpha_3$ must be $60^{\circ}$.
We will consider three cases.
(1) $\alpha_1 = 60^{\circ}$.
Then the four interior angles of the quadrilateral are $\alpha_2, 60^{\circ} + \alpha_2, \alpha_3, 60^{\circ} + \alpha_3$. Since the sum of the interior angles of the quadrilateral is $360^{\circ}$, we have $\alpha_2 + \alpha_3 = 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
\alpha_1 + \alpha_2 + \alpha_1 + \alpha_3 = 120^{\circ} + \alpha_2 + \alpha_3 = 240^{\circ} \text{.}
$$
(2) $\alpha_2 = 60^{\circ}$.

Then $\alpha_1 + \alpha_3 = 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
\alpha_1 + \alpha_3 + \alpha_3 < 2(\alpha_1 + \alpha_3) = 240^{\circ} \text{.}
$$
(3) $\alpha_3 = 60^{\circ}$.

Then $\alpha_1 + \alpha_2 = 120^{\circ}$. Therefore, $\alpha_2 < 120^{\circ}$.
Thus, the sum of the two largest interior angles of the quadrilateral is
$$
120^{\circ} + \alpha_2 < 240^{\circ} \text{.}
$$

In summary, the maximum possible value of the sum of the two largest interior angles of quadrilateral $ABCD$ is $240^{\circ}$.