Let the midpoint of $AB$ be $M$.Nadim Ul Abrar wrote:problem 16 :

In triangle $ABC$, the centroid is $G$ and $N$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $X$. Prove that $\angle AXC = \angle NGC$ if and only if angle $ACB$ is a right angle

$\angle AXC=\angle NGC \Longleftrightarrow \angle BXC=\angle BGC \Longleftrightarrow B,C,G,X$ are concyclic.

$\Longleftrightarrow BCGX$ is an isosceles trapeziod.[Since $GX \parallel BC$.]

$\Longleftrightarrow BX=CG \Longleftrightarrow \frac{1}{3}AB=\frac{2}{3}CM \Longleftrightarrow MA=MB=MC \Longleftrightarrow \angle ACB=90^{\circ}$.

Nobody solved problem $15$, so I'm not posting a new one.