The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $\pi/2$. Let $\theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $\pi-\theta$. As $\sin\theta=\sin(\pi-\theta)$, $T$ has the same area as $\triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $\triangle ABC$. — TheSimpliFire 2 mins ago