To solve this problem, let's analyze the geometry of the triangle carefully:

Given:

• $M$, $N$, and $E$ are the midpoints of $BC$, $MC$, and $AC$, respectively.
• The length of $BE$ is given as $42$ cm.
• We need to find the length of $GH$ in cm.

Step-by-Step Solution

1. Understanding Midpoint Properties and Segment Proportions: Since $M$ and $N$ are midpoints of sides $BC$ and $MC$, respectively, this configuration creates several parallel and proportional segments within the triangle $ABC$.

2. Using the Midline Theorem: The line segment $BE$ joins a vertex $B$ to the midpoint $E$ of the opposite side $AC$. In triangles, this line segment $BE$ is known as a "median," and its length is often divided proportionally when intersecting lines such as $GH$ are formed by additional midpoints within the triangle.

3. Observing Triangle Proportions in the Midline: Since $G$ and $H$ are points along the median $BE$, and we have midpoints involved, $GH$ represents a line segment within the triangle formed by these midpoints and proportional intersections. Notably, $GH$ lies parallel to $BE$ and, because of the midpoint and parallel segment properties, is exactly one-fifth the length of $BE$.

4. Calculating $GH$ Using the Given Length of $BE$: Since $BE = 42$ cm, we calculate: $GH = \frac{42}{5} = 8.4 \text{ cm}$

Thus, the correct answer is $GH = 8$ cm (rounded to the nearest whole number), not $14$ cm. The length $14$ cm would imply a larger proportional relationship that does not apply in this midpoint and parallel-segment setup.

Would you like more details on similar problems or have any further questions?

Related Questions:

1. How does the Midline Theorem apply to general triangles?
2. What are the properties of medians and midpoints in a triangle?
3. How does dividing segments proportionally help solve geometry problems?
4. Why are midpoints essential in triangle similarity problems?
5. Can you explain other applications of parallel segments within triangles?

Tip: Always look for midpoint relationships in geometry problems, as they often lead to simpler proportional calculations.