The Return Of The Cav
It’s good to see @srcav back in the twitter and blogging fold  he’s been missed!
As part of his comeback, he shared this lovely geometry puzzle:
Assuming the situation is symmetrical (which it needs to be to get a sensible solution), there are  as usual  several ways to solve it. Like Cav, I went the messy way first; however, I wanted to share a related solution using the power of complex numbers  and (much later) one that comes directly from a littleused circle theorem.
Suppose the setup is an Argand diagram

The ‘vector’ representing the left leg of the triangle corresponds to the complex number $z = 1 + \sqrt{2}i$.

Because we have a unit semicircle, the radius in the same direction is $\hat z = \frac{1}{\sqrt{3}}\br{1 + \sqrt{2}}$.

Because of the circle theorem about angles at the centre subtending double the angle at the circumference, we need to double the angle between the $x$axis and $\hat z$ to find the intersection point, which we can do by squaring $\hat z$: we get $\frac{1}{3}\br{1 + 2\sqrt{2}}$.

That puts the intersection point at $\br{\frac{1}{3}, \frac{2}{3}\sqrt{2}}$, making the width of the upper triangle a third of the big triangle.

The big triangle has area $\sqrt{2}$, so the small triangle is a ninth of that, $\frac{1}{9}\sqrt{2}$.
A circle theorem? In the wild? Someone grab it!
You might be forgiven for thinking, having heard me rant every so often, that I dislike circle theorems. That’s not true. In fact, I adore circle theorems. I just don’t think they have any place in a compulsory maths exam.
I even have a favourite! The intersecting chord theorem states that, if two chords AB and CD intersect at point P, then the products of the partchords are equal: $AP \cdot PB = CP \cdot PD$.
Interesting thing: the same result holds, even if the crossingpoint is outside the circle (and the chords are extended to become secants). It also holds even if one of the chords is, in fact, a tangent, with its endpoints in the same place.
That’s the situation pictured here: the square of the tangent $TM$ is equal to the product $LM \cdot XM$.
Now, $TM$ takes a tiny bit of thought to figure out: triangle OTM is clearly rightangled at T; its hypotenuse is $\sqrt{2}$ and the leg $OT$ is 1  so $TM=1$.
We know $LM$ from Pythagoras; it’s $\sqrt{3}$. The only thing we don’t know is $XM$ (let’s call that distance $x$). We have $\sqrt{3} x = 1$, so $x = \frac{\sqrt{3}}{3}$.
That means $XM$ is a third of the length of $LM$, and since triangle $LMR$ is similar to $XMX’$, the smaller triangle has a ninth of the area of the large one.
I think that’s a really elegant approach  but I’d be interested to see anything else that’s as neat!
* Many thanks to @srcav and @profsmudge for discussions on this problem.