The problem describes a triangle $ABC$ with a parallel line segment $EF$ inside, where:

• $AC = 36 \, \text{cm}$ (the base of the triangle),
• $BC = 30 \, \text{cm}$ (another side of the triangle),
• $EF = 12 \, \text{cm}$ (the parallel segment),
• $BE = 9 \, \text{cm}$ (the part of side $BC$).

We are tasked with finding the lengths of $AB$ and $BF$.

This setup indicates that triangle $ABC$ is similar to triangle $BEF$ because the line segment $EF$ is parallel to side $AC$.

Step 1: Set up the ratios for similar triangles

Since triangles $ABC$ and $BEF$ are similar by the property of parallel lines, the sides are proportional: $\frac{BE}{BC} = \frac{EF}{AC}$

Substitute the known values: $\frac{9}{30} = \frac{12}{36}$

Both ratios simplify to $\frac{3}{10}$, confirming that the triangles are indeed similar.

Step 2: Calculate the unknown lengths

Using the similarity of the triangles, the ratio of corresponding sides is $\frac{3}{10}$.

Finding $AB$:

From the similarity, we can find $AB$ using the ratio: $\frac{AB}{AC} = \frac{BE}{BC}$ $\frac{AB}{36} = \frac{3}{10}$ Solving for $AB$: $AB = \frac{3}{10} \times 36 = 10.8 \, \text{cm}$

Finding $BF$:

We know that $BF = BC - BE$, so: $BF = 30 - 9 = 21 \, \text{cm}$

Final answers:

• $AB = 10.8 \, \text{cm}$
• $BF = 21 \, \text{cm}$

Would you like further details or have any questions? Here are five related questions to deepen your understanding:

1. How do you know that triangles $ABC$ and $BEF$ are similar?
2. What other properties can be derived from the similarity of these triangles?
3. Can you calculate the height of triangle $ABC$ using known values?
4. What happens to the similarity ratio if the parallel segment $EF$ is moved closer to the base $AC$?
5. How would the problem change if triangle $ABC$ were isosceles?

Tip: Always check if triangles share proportional sides or angles to determine similarity.