The problem provides a diagram with $AEXD$ and $BFXC$ being straight lines. We are given the lengths $AE = 2$ cm, $EX = 4$ cm, $XD = 6$ cm, and $CD = 7\frac{1}{2}$ cm. Also, $AB$, $EF$, and $CD$ are parallel. We need to find a triangle congruent to $\triangle XCD$, a triangle similar but not congruent to $\triangle XCD$, and the length of $EF$.

1. Problem Description

The problem provides a diagram with $AEXD$ and $BFXC$ being straight lines. We are given the lengths $AE = 2$ cm, $EX = 4$ cm, $XD = 6$ cm, and $CD = 7\frac{1}{2}$ cm. Also, $AB$, $EF$, and $CD$ are parallel. We need to find a triangle congruent to $\triangle XCD$, a triangle similar but not congruent to $\triangle XCD$, and the length of $EF$.

2. Solution Steps

a) Finding congruent and similar triangles:

Since $AB$, $EF$, and $CD$ are parallel, we can observe similar triangles.

$\triangle AEX \sim \triangle DCX$.

$\triangle BEX \sim \triangle FCX$.

$\triangle ABX \sim \triangle CDX$.

i) Congruent to $\triangle XCD$: It seems like the question expects that only similar triangles will be considered and it is not obvious whether there is a congruent triangle. Given the information in the problem statement, we can't prove that $\triangle ABX \cong \triangle CDX$. So it's possible the question is designed such that a trivial answer $\triangle XCD$ is expected.

ii) Similar but not congruent to $\triangle XCD$:

Since $AB$, $EF$, and $CD$ are parallel, $\triangle ABX$ is similar to $\triangle EFX$ and $\triangle CDX$.

Since $AE$ and $XD$ have different lengths, $AB$ and $CD$ have different lengths, thus $\triangle ABX$ and $\triangle CDX$ are not congruent.

Therefore, $\triangle ABX$ is similar to $\triangle XCD$, but not congruent to $\triangle XCD$.

b) Finding the length of $EF$:

Since $AB$, $EF$, and $CD$ are parallel, we can use similar triangles to find the length of $EF$.

We have that $\triangle EFX \sim \triangle CDX$.

Thus, we have the proportion $\frac{EF}{CD} = \frac{EX}{XD}$.

We are given that $EX = 4$ cm, $XD = 6$ cm, and $CD = 7\frac{1}{2} = \frac{15}{2}$ cm.

So, we have $\frac{EF}{\frac{15}{2}} = \frac{4}{6} = \frac{2}{3}$.

$EF = \frac{2}{3} \cdot \frac{15}{2} = 5$ cm.

3. Final Answer

a) i) $\triangle XCD$

ii) $\triangle ABX$

b) $EF = 5$ cm