In the given figure, EF ext || DG and AF ext || BG. If DC = 33 cm, ED = 22 cm, and BC = 15 cm, find the length of AB.

In $\triangle CEF$, we are given that $EF \parallel DG$. This means that $\triangle CDG$ is similar to $\triangle CFE$ (by AA similarity, as $\angle C$ is common and $\angle CDG = \angle CFE$ due to corresponding angles).

From the similarity $\triangle CDG \sim \triangle CFE$, we have the ratio of corresponding sides:

We are given $ED = 22 \text{cm}$ and $DC = 33 \text{cm}$.
So, $CE = CD + DE = 33 \text{cm} + 22 \text{cm} = 55 \text{cm}$.
Now, substitute the values into the ratio:

In $\triangle ACF$, we are given that $AF \parallel BG$. This means that $\triangle CBG$ is similar to $\triangle CAF$ (by AA similarity, as $\angle C$ is common and $\angle CBG = \angle CAF$ due to corresponding angles).

From the similarity $\triangle CBG \sim \triangle CAF$, we have the ratio of corresponding sides:

From Step 2, we know $\frac{CG}{CF} = \frac{3}{5}$.
So,

We are given $BC = 15 \text{cm}$ and $AB = x$.
Therefore, $CA = CB + AB = 15 \text{cm} + x$.
Substitute these values into the ratio:

Multiply both sides by $5(15+x)$:

Subtract 45 from both sides:

Divide by 3:

The value of $x$ is $10 \text{cm}$.