Rijul saini wrote: Let $M$ be an interior point of a $\triangle ABC$ such that $\angle AM B = 150^{\circ} , \angle BM C = 120^{\circ}$. Let $P, Q, R$ be the circumcentres of the $\triangle AM B, \triangle BM C, \triangle CM A$ respectively. Prove that $[P QR] \ge [ABC]$. For this, we just recognise that $\triangle PQR$ is right angled. The rest is quite easy now

Recognizing that $ \triangle PQR $ is a right is easy. How do you know the rest is easy?

The inequality is still true even if $ \triangle PQR $ is not a right triangle. $ \triangle PQR $ is right triangle just makes it easy for angle computation.