The triangle must be obtuse if two of its medians are perpendicular? In problem #577679, the question says if two of the medians of triangle ABC are perpendicular, then .... ($5a^2 = b^2 + c^2$, the result).

In the course of solving it, I drew the following picture (slightly different to that question).


I found that angle ABC 'must be obtuse' before I can get a satisfactory (hand) drawing. (I did not use any tool from Wolfram.)

My statement is "Prove or disprove the triangle must be obtuse under the mentioned given".

I tried to use the cosine law to show that $cos ABC$ (or $a^2 + c^2 - b^2$) is negative, but was not successful.

Any idea?
(This old question came up again through an edit by @JonMarkPerry.)

Let's find all acute triangles with two perpendicular medians. We may assume the following situation (see the figure):
$$A=(0,0), \quad B=(c,0),\quad C=(u,v), \quad M_c=\left({c\over2},0\right), \quad M_b=\left({u\over2},{v\over2}\right)$$
with $v&gt;0$. We want that $$BM_b\perp CM_c\ ,\tag{1}$$ which means that
$$\overrightarrow{BM_b}\cdot\overrightarrow{CM_c}=\left({u\over2}-c,{v\over2}\right)\cdot\left({c\over2}-u,-v\right)=0$$
or
$$\left(u-{5\over4}c\right)^2+ v^2={9\over16}c^2\ .$$
It follows that the points $C$ satisfying $(1)$ are lying on the circle $K$ with center $\bigl({5\over4}c,0\bigr)$ and radius ${3\over4}c$. The triangle  $ABC$ is acute iff $C$ lies outside of the Thales circle over $AB$ and to the left of the line $u=c$. This part of $K$ is drawn in bold red in the figure.
Going down the route of using cosine rule, if all the angles are less than $90$ then $c^2+a^2&gt;b^2, c^2+b^2&gt;a^2, b^2+a^2&gt;c^2$. Additionally, $c^2+b^2=5a^2$. Hence one gets the inequalities $(6b^2)/5&gt;(4c^2)/5/&gt;(8b^2)/15$ and a similar one where $b$ and $c$ are interchanged. As an example, with $b=1, c=0.9, a=\frac{\sqrt{905}}{50}$, all the inequalities, including the triangle inequality are satisfied, providing an example of medians perpendicular to each other without being obtuse.