Question

Plane Geometry
Find the acute, angle between the medians of an isosceles right triangle which are drawn from the vertices of its acute angles.

Answer 1

From figure ; $\angle BAC=\angle BCA={45}^0$

$CE$ & $AD$ bisect $AB$ & $BC$ resp.

$\Rightarrow BE=BD=\frac{x}{2}$

Now, In $\Delta ABD$,

${AD}^2={AB}^2+{BD}^2=x^2+\frac{x^2}{4}$

$\Rightarrow AD=\frac{x\sqrt{5}}{2}$

Also, $AO=CO=\frac{2}{3}\left(AD\right)=\frac{2}{3}\times \frac{x\sqrt{5}}{2}=\frac{x\sqrt{5}}{3}$

Now, $\angle AOC=\frac{{AO}^2+{CO}^2-{AC}^2}{2AO.CO}$

$\cos \angle AOC=\frac{2\left({5x}^2/9\right)-\left({2x}^2\right)}{2\left({5x}^2/9\right)}=1-\frac{9}{5}=-\frac{4}{5}$

$\angle AOC={\cos }^{-1}(-4/5)={36.87}^0$