Here is an alternate solution to problem 36We apply cartesian coordinates.
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Note that the problem is equivalent to proving $F',D,E$ collinear in the diagram.
Let $U\equiv (a,1), V\equiv (-a,1), T\equiv (at,t), S\equiv (-as,s)$.
Now, let $E \equiv (p,q)$. We get, $\dfrac{q-t}{p-at}=-a$ and $\dfrac{q-s}{p+as}=a$. From these, we get $E \equiv (\dfrac{(a^2+1)(t-s)}{2a},\dfrac{(a^2+1)(t+s)}{2})$.
Again, $D$ lies on the $y$-axis. From $DU\perp OU$, we get that $D\equiv (a^2+1,0)$.
Now for the *most difficult* part, the coordinates of $F$. The slope of $ST$ is $\dfrac{t-s}{a(t+s)}$. So the slope of $OF$ is $\dfrac{a(t+s)}{s-t}$. So for some real $m$, $F\equiv (m(s-t),ma(s+t))$.
Now, $FS\perp OT$, so $\dfrac{ma(s+t)-s}{m(s-t)+as}=-a$. So we get $m=\dfrac{1-a^2}{2a}$. So $F\equiv (\dfrac{(1-a^2)(s-t)}{2a},\dfrac{(1-a^2)(s+t)}{2})$. So $F'\equiv (\dfrac{(1-a^2)(s-t)}{2a},2-\dfrac{(1-a^2)(s+t)}{2})$.
Now, apply the shoelace formula to $F',D,E$. The det is $0$, so they are collinear.
Time taken: 30 minutes while simultaneously taking english notes.
Hammer with tact.
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