Geometry question: Find the area of blue-shared area inside this isosceles See below, looks a bit interesting, but I cannot find a solution.
I think a starting point might be the similarity of the lower white triangle and the larger triangle composed of lower blue, pink, and lower right triangle though..

 A: I've found a solution without trig, but always leading to a cubic equation.
Notice first of all that if you have a square inscribed into a right triangle, with the right angle in common, then the side of the square is equal to the product of the legs of the triangle, divided by their sum (this can be easily proved using triangle similarity).
Let now $a=AD$, $b=BD$ and $c=CD$ (see figure below). By the above observation we get then two equations:
$$
ab=7(a+b),\quad bc=16(b+c).
$$
As $AC=BC$ we also have:
$$
(a+c)^2=b^2+c^2,
\quad\text{that is:}\quad
2ac=b^2-a^2.
$$
From the last equation we get:
$$
c={b^2-a^2\over2a}.
$$
Plugging that into the second equation we obtain:
$$
(b-16)(b^2-a^2)=32ab=32\cdot7(a+b)
$$
and we can factor out $(a+b)$ to find
$$
a=b-{224\over b-16}.
$$
Inserting that into the first equation finally leads to a cubic equation for $b$:
$$
b^3-30b^2+1568=0.
$$
Trying the divisors of $1568=2^5\cdot7^2$ greater than $16$ we can find that $b=28$ is a solution: factoring leads then to a quadratic equation whose solutions can be discarded (one is negative and the other is less than $16$).
Hence $b=28$ is the only solution and from that the requested area can be easily computed.

A: Expanding-upon a comment, as requested.

Let the triangle have vertex angle $2\theta$, let $a\leq b$ be the sides of the squares, and let $t$ denote the target area. We immediately find
$$\begin{align}
a+a\cot\theta = |PQ| = b+b\tan2\theta \tag{1} \\
t = \frac12\cdot a\cdot a\cot\theta + \frac12\cdot b \cdot b\tan2\theta \tag{2}
\end{align}$$
From here, "all we have to do" is eliminate $\theta$ from $(1)$ and $(2)$. One way to do that here is first to treat the equations as a linear system in $\cot\theta$ and $\tan2\theta$; solving gives
$$\cot\theta = \frac{2t-b(a-b)}{a (a + b)} \qquad 
\tan2\theta = \frac{2t+a(a-b)}{b (a + b)} \tag{3}$$
Reciprocating $\cot\theta$ into $\tan\theta$, and substituting into the double-angle identity $\tan2\theta=2\tan\theta/(1-\tan^2\theta)$, and "simplifying" gives this relation among $t$, $a$, and $b$:
$$\begin{align}
0 
&= 8 t^3 \\
&+ 4 t^2 (a - b)(a - 2 b)  \\
&- 2 t\phantom{^2}(a^2 + b^2)(a^2 + 6 a b - b^2)  \\
&- \phantom{2}a (a - b) (a^4 + 2 a^3 b - 2 a^2 b^2 - 2 a b^3 - 3 b^4) 
\end{align} \tag{$\star$}$$
In the particular case of $a=7$ and $b=16$, $(\star)$ reduces into a factorable form and we have
$$(2t-339) (4t^2+1128t+49371) = 0 \quad\to\quad t = \frac{339}{2} \tag{$\star$'}$$
which gives our answer. $\square$

As an alternative to the above approach with $(3)$ we could re-write $(1)$  in terms of $\tan\theta$ alone:
$$(a-b)\tan^3\theta + (a+2b)\tan^2\theta-(a-b)\tan\theta-a=0 \tag{4}$$
For $a=7$ and $b=16$, this reduces to
$$(3\tan\theta-1) (3\tan^2\theta-12\tan\theta-7) =0 \tag{5}$$
Therefore (casually ignoring a positive root of the second factor),
$$\tan\theta = \frac13\qquad\to\qquad\tan2\theta=\frac{3}{4}\qquad\stackrel{\text{by (1)}}{\to}\qquad t = \frac{339}{2}\tag{6}$$
(See the edit history that also rewrites $(2)$ in terms of $\tan\theta$ alone before eliminating $\tan\theta$ from the system to get $(\star)$.)
An advantage of this approach is that $(4)$ is "more-immediate" to factor. Also, we see explicitly that $\tan\theta$ and $\tan2\theta$ are nice values. Knowing $\tan\theta=1/3$ facilitates drawing an accurate figure.
More importantly, $\tan2\theta=3/4$ reveals a non-obvious connection to the $3$-$4$-$5$ triangle. I can't help thinking that there's a clever way to detect and exploit it.

Yet-another way to proceed is to seek the distance $p:=|PQ|$.  (@Aretino's answer uses another approach to this.) Clearly,
$$\tan\theta = \frac{a}{p-a} \quad \tan2\theta=\frac{p-b}{b} \qquad\stackrel{\text{dbl ang}}{\to}\qquad p^3 - p^2(2 a + b) + 2 a^2 b  =0 \tag{7}$$
(The nicest cubic yet!) For $a=7$ and $b=16$, this reduces to
$$(p-28) (p^2-2p-56)=0 \quad\to\quad p = 28 \tag{8}$$
from which the result follows thusly:
$$t = \frac12a(p-a)+\frac12b(p-b) = \cdots = \frac{339}{2} \tag{9}$$
This solution seems like it might be the most accessible to middle-schoolers, although it, too, ignores the underlying $3$-$4$-$5$ triangle. (Perhaps I'm over-estimating the significance of that triangle!)
