Let $ABC$ be a triangle. Let the $A$-altitude meet $BC$ at $D$, the $B$-altitude meet $AC$ at $E$, and let $T \neq A$ be the point on the circumcircle of $\triangle ABC$ such that $AT \parallel BC$. Given that $D, E, T$ are collinear, if $BD = 3$ and $AD = 4$, then the area of $\triangle ABC$ can be written as $a + \sqrt{b}$, where $a$ and $b$ are positive integers. What is $a + b$?

Let's think step by step and output the final answer within \boxed{}.
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