We are given a triangle with some line segments inside it. We are given the lengths $OH = 9$ cm, $HA = 5$ cm, $HK = 6$ cm, and $AB = 8$ cm. We are asked to calculate the lengths of $OK$ and $KB$. We assume that triangle $OHA$ is similar to triangle $OKB$.

1. Problem Description

We are given a triangle with some line segments inside it. We are given the lengths $OH = 9$ cm, $HA = 5$ cm, $HK = 6$ cm, and $AB = 8$ cm. We are asked to calculate the lengths of $OK$ and $KB$. We assume that triangle $OHA$ is similar to triangle $OKB$.

2. Solution Steps

Since triangles $OHA$ and $OKB$ are similar, we have the following proportion:

$\frac{OH}{OK} = \frac{HA}{KB} = \frac{OA}{OB}$.

We know that $OA = OH + HA = 9 + 5 = 14$ cm.

Also $OK = OH + HK = 9 + 6 = 15$ cm.

Using the similarity of the triangles, we have

$\frac{OH}{OK} = \frac{9}{15} = \frac{3}{5}$ and $\frac{HA}{KB} = \frac{5}{KB}$.

Since $\frac{OH}{OK} = \frac{HA}{KB}$, we have $\frac{3}{5} = \frac{5}{KB}$.

Therefore, $3 \cdot KB = 5 \cdot 5 = 25$, so $KB = \frac{25}{3}$.

Also, $\frac{OA}{OB} = \frac{14}{AB+AK}$

Also, given that $AB=8$ cm.

Since $OK = OH + HK$, we have $OK = 9 + 6 = 15$ cm.

From the similarity $\frac{OH}{OK} = \frac{HA}{KB}$, we get

$\frac{9}{15} = \frac{5}{KB}$.

$9KB = 15 \cdot 5 = 75$.

$KB = \frac{75}{9} = \frac{25}{3}$ cm.

3. Final Answer

a) $OK = 15$ cm

b) $KB = \frac{25}{3}$ cm