How to algebraically demonstrate that the statement below is false? A triangle always occupies more than a quarter of the area of the circle it circumscribes.
Visually, it is easy to see that the statement is false, but I would like an algebraic demonstration using plane geometry
I try:
$b=2R$
$S_\triangle ABC = \frac{2R.h}{2}=R.h$
$S\triangle ABC: =\dfrac{abc}{4R} < \dfrac{\pi R^2}{4} \implies \dfrac{ac}{2} < \dfrac{\pi R^2}{4}$
$Rh=\dfrac{ac}{2}\implies ac = 2Rh \implies Rh < \dfrac{\pi R^2}{4} $
???
As indicated in your diagram, consider the right-angled $\triangle ABC$ where $\measuredangle ABC = 90^{\circ}$, so $AC$ is the hypotenuse, and with $\lvert AC\rvert = R$. Its circumscribed circle $\Large\circ$ then has a diameter of $R$.
Let $0 \lt m \le 0.5$ be the multiplier of $R$ that is the height of the triangle from $AC$, i.e., we have
$$h = mR$$
Then
$$\left(\frac{S_{\triangle ABC}}{S_{\Large\circ}} = \frac{\left(\frac{(mR)R}{2}\right)}{\left(\frac{\pi R^2}{4}\right)}\right) \le \frac{1}{4} \iff \frac{2m}{\pi} \le \frac{1}{4} \iff m \le \left(\frac{\pi}{8} \approx 0.3927\right)$$
gives the set of values of $m$ which provide counter-examples to