Problem 1313

1. Assume the given conditions:
- We have a triangle $\triangle ABC$.
- Three cevians $AD, BE,$ and $CF$ are drawn from vertices $A, B,$ and $C$ respectively, intersecting at a common point $G$.
- These cevians divide the triangle into 4 smaller triangles and 3 quadrilaterals.

2. Identify the quadrilaterals:
- The quadrilaterals formed are $DGHB, EHIC,$ and $AGIF$.

3. Assume that all three quadrilaterals are cyclic:
- A quadrilateral is cyclic if its opposite angles sum to $180^\circ$.

4. Analyze the cyclic quadrilaterals:
- For $DGHB$ to be cyclic, $\angle DGH + \angle DBH = 180^\circ$.
- For $EHIC$ to be cyclic, $\angle EHI + \angle ECI = 180^\circ$.
- For $AGIF$ to be cyclic, $\angle AGI + \angle AFI = 180^\circ$.

5. Consider the implications for the triangles:
- If $DGHB$ is cyclic, then $\angle DGH = \angle DBH$.
- If $EHIC$ is cyclic, then $\angle EHI = \angle ECI$.
- If $AGIF$ is cyclic, then $\angle AGI = \angle AFI$.

6. **Prove that the triangles $AEC, BDC,$ and $ABF$ are isosceles:**
- Since $DGHB$ is cyclic, $\angle DGH = \angle DBH$, implying $\triangle BDC$ is isosceles with $BD = DC$.
- Since $EHIC$ is cyclic, $\angle EHI = \angle ECI$, implying $\triangle AEC$ is isosceles with $AE = EC$.
- Since $AGIF$ is cyclic, $\angle AGI = \angle AFI$, implying $\triangle ABF$ is isosceles with $AF = FB$.

7. Derive a contradiction:
- From the isosceles triangles, we have $AC = CE$, $BC = BD$, and $AB = AF$.
- This implies $AC = CE < BC = BD < AB = AF < AC$, which is a contradiction because it implies a cyclic inequality that cannot hold true.

8. Conclude that the assumption is false:
- Since assuming all three quadrilaterals are cyclic leads to a contradiction, it is not possible for all three quadrilaterals to be inscribed.

$\blacksquare$

The final answer is False.