Given triangle $OFC$, points $A$ and $B$ are on segment $OC$ such that $A$ is closer to $O$ and $B$ is closer to $C$. Line through $A$ parallel to $BF$ intersects $OF$ at $E$. Line through $B$ parallel to $CE$ intersects $OF$ at $D$. The problem asks us to: 1. Draw the figure.

1. Problem Description

Given triangle $OFC$, points $A$ and $B$ are on segment $OC$ such that $A$ is closer to $O$ and $B$ is closer to $C$. Line through $A$ parallel to $BF$ intersects $OF$ at $E$. Line through $B$ parallel to $CE$ intersects $OF$ at $D$.

The problem asks us to:

1. Draw the figure.

2. Show that there exists a homothety $h_1$ transforming $A$ to $B$ and $E$ to $F$.

3. Characterize the homothety $h_1$.

4. Show that there exists a homothety $h_2$ transforming $B$ to $C$ and $D$ to $E$.

5. Characterize the homothety $h_2$.

6. Characterize the application $f = h_2 \circ h_1$. Justify that $h_2 \circ h_1 = h_1 \circ h_2$.

7. Determine $f(A)$ and $f(D)$ and deduce that $(AD)$ and $(CF)$ are parallel.

2. Solution Steps

1. Figure Description:

Draw triangle $OFC$. Place points $A$ and $B$ on segment $OC$ such that $A$ is closer to $O$ than $B$ is to $C$. Draw a line through $A$ parallel to $BF$, intersecting $OF$ at $E$. Draw a line through $B$ parallel to $CE$, intersecting $OF$ at $D$.

2. Existence of $h_1$:

Since $AE \parallel BF$, by Thales' theorem in triangle $OFC$: $\frac{OA}{OC} = \frac{OE}{OF}$.

We want to find a homothety $h_1$ such that $h_1(A) = B$ and $h_1(E) = F$. A homothety is defined by its center and ratio.

If such a homothety exists, then $A, B$, and the center must be collinear, and $E, F$, and the center must be collinear. Let the center be $X$. Then $X$ must lie on both $OC$ and $OF$, so $X = O$.

Then we need $\vec{OB} = k \vec{OA}$ and $\vec{OF} = k \vec{OE}$. So $k = \frac{OB}{OA}$ and $k = \frac{OF}{OE}$. Therefore, $\frac{OB}{OA} = \frac{OF}{OE}$.

Since $AE \parallel BF$, by Thales' theorem $\frac{OE}{OF} = \frac{OA}{OB}$.

So, $\frac{OF}{OE} = \frac{OB}{OA}$. This equality shows the existence of the homothety $h_1$.

3. Characterizing $h_1$:

The center of $h_1$ is $O$ and the ratio is $k_1 = \frac{OB}{OA} = \frac{OF}{OE}$.

4. Existence of $h_2$:

We want to find a homothety $h_2$ such that $h_2(B) = C$ and $h_2(D) = E$.

If such a homothety exists, then $B, C$, and the center must be collinear, and $D, E$, and the center must be collinear. Let the center be $Y$. Then $Y$ must lie on both $OC$ and $OF$, so $Y=O$.

Then we need $\vec{OC} = k' \vec{OB}$ and $\vec{OE} = k' \vec{OD}$. So $k' = \frac{OC}{OB}$ and $k' = \frac{OE}{OD}$. Therefore, $\frac{OC}{OB} = \frac{OE}{OD}$.

Since $BD \parallel CE$, by Thales' theorem $\frac{OD}{OE} = \frac{OB}{OC}$.

So, $\frac{OE}{OD} = \frac{OC}{OB}$. This equality shows the existence of the homothety $h_2$.

5. Characterizing $h_2$:

The center of $h_2$ is $O$ and the ratio is $k_2 = \frac{OC}{OB} = \frac{OE}{OD}$.

6. Characterizing $f$ and $h_2\circ h_1 = h_1 \circ h_2$:

$f = h_2 \circ h_1$ is a homothety with center $O$ and ratio $k_2k_1 = \frac{OC}{OB} \frac{OB}{OA} = \frac{OC}{OA}$. Since the center of $h_1$ and $h_2$ are the same point $O$, then $h_2 \circ h_1 = h_1 \circ h_2$.

7. $f(A)$ and $f(D)$ and $(AD) \parallel (CF)$:

$f(A) = h_2(h_1(A)) = h_2(B) = C$.

$f(D) = h_2(h_1(D))$. Since $h_1(D)$ is a point $X$ on the line $OD$ such that $\frac{OX}{OD} = \frac{OB}{OA}$. Thus $OX = \frac{OB}{OA}OD$. Then $f(D) = h_2(h_1(D)) = h_2(X) = Y$ on the line $OX=OD$ such that $OY = \frac{OC}{OB}OX = \frac{OC}{OB} \frac{OB}{OA} OD = \frac{OC}{OA} OD$.

Since $f(A) = C$ and $f(D) = Y$, where $Y$ is on $OD$ and $\frac{OY}{OD} = \frac{OC}{OA}$.

In triangle $ODA$, we have $\frac{OA}{OC} = \frac{OD}{OY}$. Therefore, by the converse of Thales' Theorem, $AD \parallel CY$ which is $CF$.

3. Final Answer

1. Figure: Described above.

2. Homothety $h_1$ exists.

3. $h_1$ has center $O$ and ratio $\frac{OB}{OA}$.

4. Homothety $h_2$ exists.

5. $h_2$ has center $O$ and ratio $\frac{OC}{OB}$.

6. $f$ is a homothety with center $O$ and ratio $\frac{OC}{OA}$. $h_2 \circ h_1 = h_1 \circ h_2$ because they have the same center.

7. $f(A) = C$ and $f(D)$ lies on the line $OD$ such that $\frac{OF(D)}{OD} = \frac{OC}{OA}$. Therefore, $AD \parallel CF$.