Prove that AC2 = AD AE and EF AC = ED * AF

Let's break down this geometry problem.

First, we need to identify similar triangles from the given diagram.
From the diagram, we observe:

• The line segment $FC$ is perpendicular to the line segment $CD$ at point $C$, so $\angle FCA = 90^\circ$.
• The line segment $ED$ is perpendicular to the line segment $CD$ at point $D$, so $\angle EDA = 90^\circ$.
• The lines $FE$ and $CD$ intersect at point $A$. Therefore, $\angle FAC$ and $\angle EAD$ are vertically opposite angles, which means $\angle FAC = \angle EAD$.

Since two angles of $\triangle FAC$ are equal to two angles of $\triangle EAD$, the triangles are similar by the Angle-Angle (AA) similarity criterion.
Thus, $\triangle FAC \sim \triangle EAD$.

From the similarity of these triangles, the ratio of their corresponding sides is equal:

$$\frac{FC}{ED} = \frac{AC}{AD} = \frac{FA}{EA}$$

Now let's address each part of the question.

Part 1: Prove that $EF \cdot AC = ED \cdot AF$

From the similarity ratio $\frac{AC}{AD} = \frac{FA}{EA}$, we can write $AC = \frac{FA \cdot AD}{EA}$.
Also, from the diagram, points $F, A, E$ are collinear, and $A$ is between $F$ and $E$. So, $EF = FA + AE$.

Let's substitute these into the equation we need to prove:

$$(FA + AE) \cdot AC = ED \cdot AF$$

Substitute $AC$:

$$(FA + AE) \cdot \left(\frac{FA \cdot AD}{EA}\right) = ED \cdot AF$$ $$\frac{FA^2 \cdot AD}{EA} + \frac{AE \cdot FA \cdot AD}{EA} = ED \cdot AF$$ $$\frac{FA^2 \cdot AD}{EA} + FA \cdot AD = ED \cdot AF$$

Factor out $FA \cdot AD$:

$$FA \cdot AD \left(\frac{FA}{EA} + 1\right) = ED \cdot AF$$ $$FA \cdot AD \left(\frac{FA + EA}{EA}\right) = ED \cdot AF$$

Since $FA + EA = EF$:

$$FA \cdot AD \cdot \frac{EF}{EA} = ED \cdot AF$$

Divide both sides by $AF$ (assuming $AF \neq 0$):

$$AD \cdot \frac{EF}{EA} = ED$$ $$AD \cdot EF = ED \cdot EA$$

From the similarity ratio $\frac{FC}{ED} = \frac{FA}{EA}$, we can write $ED = \frac{FC \cdot EA}{FA}$.
Substitute this into the equation:

$$AD \cdot EF = \left(\frac{FC \cdot EA}{FA}\right) \cdot EA$$ $$AD \cdot EF = \frac{FC \cdot EA^2}{FA}$$

This expression is not generally true and does not simplify to an identity. This indicates that the statement $EF \cdot AC = ED \cdot AF$ is not generally true based on the given similar triangles and collinearity. It is highly probable there is a typo in the question. A common relation derived from these similar triangles is $FC \cdot AD = ED \cdot AC$.

Part 2: Prove that $AC^2 = FC \cdot AD$

From the similarity ratio $\frac{FC}{ED} = \frac{AC}{AD}$, we can cross-multiply to get:

$$AC \cdot ED = FC \cdot AD$$

For the statement $AC^2 = FC \cdot AD$ to be true, we would need $AC^2 = AC \cdot ED$, which implies $AC = ED$.
However, $AC$ is a segment on the line $CD$, and $ED$ is a perpendicular segment. They are not necessarily equal. Therefore, the statement $AC^2 = FC \cdot AD$ is also not generally true based on the given similar triangles.

Given that both statements are not generally true under the standard interpretation of similar triangles, it is likely there are typos in the question. If the question intended to ask for relations directly derived from the similarity, they would be:

1. $FC \cdot AD = ED \cdot AC$ (from $\frac{FC}{ED} = \frac{AC}{AD}$)
2. $FA \cdot AD = EA \cdot AC$ (from $\frac{AC}{AD} = \frac{FA}{EA}$)
3. $FC \cdot EA = ED \cdot FA$ (from $\frac{FC}{ED} = \frac{FA}{EA}$)

Since I must answer the question as written, and the statements are not generally true, I will state the conclusion.

Conclusion:
Based on the similarity $\triangle FAC \sim \triangle EAD$, the given statements are not generally true.

1. The statement $EF \cdot AC = ED \cdot AF$ implies $AD \cdot EF = ED \cdot EA$, which is not generally true.
2. The statement $AC^2 = FC \cdot AD$ implies $AC = ED$, which is not generally true.

It is possible there is a typo in the question.