I realize that some of this is going to come off as pretentious and I'm sorry if so. I'm just genuinely curious. Recently I've made an observation regarding isosceles triangles.
Take some number $\omega$ and $\triangle ABC$, where $C$ is the top vertex. If $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ both have a length of $\omega \sqrt5$, and $\overleftrightarrow{AB}$ has a length of $2\omega$, then there exists an inscribed square $\square LMNO$ with sides of length $\omega$ where two of the vertices are the midpoints of $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ respectively. Here is an example $\triangle ABC$ to illustrate what I mean.
I don't think this is life-changing or anything, I just think it's really cool and would like some help writing a proof for it.
Let us denote the midpoint of $AB$ (which is the origin in your diagram) by $D$.
Define $L, M, N, O$ to be the midpoints of $DA, AC, CB, BD$ respectively. We need to show that $LMNO$ is indeed a square with side length $\omega$.
By midline/midpoint theorem, we have $ML // CD$ and $ML = \frac12CD$.
Similarly $NO // CD$ and $NO = \frac12CD$.
By the properties of an isosceles triangle (or SSS congruence), $AB \perp CD$.
By Pythagoras Theorem, $CD = \sqrt{AC^2 - AD^2} = \sqrt{5\omega^2 - \omega^2} = 2\omega$.
Hence $ML = NO = \omega$ and $ML // NO$. This shows that $LMNO$ is a parallelogram.
Since $LO = 2(\frac12\omega) = \omega = ML = NO$, $LMNO$ is a rhombus.
Since $LM // CD$, $LM \perp AB$. Hence $LMNO$ is a square.