The answer is yes. Here's the details: Denote a side triangle as a triangle which shares the vertex with the polygon and it has at least one side coincides with the polygon. Claim 01. Suppose that the polygon itself has an acute interior angle. Then the polygon clearly satisfies the problem condition. Proof. Take that particular vertex that has an acute interior angle. Any diagonal will do the job, as it clearly divides the acute angle into two acute angles. Claim 02. $n \ge 5$. Proof. This clearly follows from the fact that From Claim 01, the case $n = 4$ melts to the case of it being a rectangle. But this clearly satisfies the problem condition, as any right angle must be divided into two acute angles. For Claim 03 and onwards, denote right angle and obtuse as one group: obtuse itself. Claim 03. All polygons with existing acute side triangle satisfy. Proof. Let the polygon itself be $A_1A_2A_3 \dots A_n$. Suppose there exists an acute side triangle, WLOG let it be $A_1 A_2 A_i$. Therefore, considering diagonal $A_2A_i$ and vertex $A_2$, this gives us that $\angle A_i A_2 A_3$ must be obtuse. Continuing this onwards, we have that $\angle A_i A_n A_{n+1}$ must be obtuse for every $n \in \mathbb{N}$. Meanwhile, this gives us $\angle A_iA_{i - 2}A_{i - 1}$ being obtuse. Similarly, do it from the other way round: Considering diagonal $A_1A_i$ and vertex $A_1$, this gives us that $\angle A_n A_1 A_i$ must be obtuse. This gives us that $\angle A_nA_{n+1} A_i$ must be obtuse for every $n \in \mathbb{N}$. So, we will have $\angle A_{i + 1} A_{i + 2}A_i$ must be obtuse. Now, notice that from the above two fact, we have now that $\angle A_{i + 2}A_{i + 1}A_i$ must be acute. By Claim 01, we are hence finished. Claim 04. All polygons satisfy the question. Proof. From the previous claim, we are thus left with the case dealing with no acute triangles left. In this case, all of the side triangles were obtuse. Now, notice that by a similar method as in Claim 03, one can force an angle of the polygon to be acute later, and this finishes the problem.

Dadgarnia wrote: Is it true that in any convex $n$-gon with $n > 3$, there exists a vertex and a diagonal passing through this vertex such that the angles of this diagonal with both sides adjacent to this vertex are acute? Proposed by Boris Frenkin - Russia Just take the line segment of maximal length determined by the vertices of the polygon, if it is a side of the polygon, then at least two of the angles of the polygon must be acute, so in this case there's nothing to do. If the maximal length line segment is a diagonal then choosing any of the angles at the endpoints of the diagonal works. Here we are using the simple fact that in an obtuse triangle the side opposite to the obtuse angle is the longest.

@above yup, that’s how I solved it too =P

Jajaja I solve it just by induction...