If OF = 2/5 (a+b), express DE in terms of a and b.

If F is on the line segment AD, then $\vec{OF}$ can be expressed as a linear combination of $\vec{OA}$ and $\vec{OD}$ using the section formula:

$$\vec{OF} = (1-t)\vec{OA} + t\vec{OD}$$ $$\vec{OF} = (1-t)\vec{a} + t\vec{b}$$

We are given $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b}) = \frac{2}{5}\vec{a} + \frac{2}{5}\vec{b}$.
Comparing the coefficients of $\vec{a}$ and $\vec{b}$ (assuming $\vec{a}$ and $\vec{b}$ are not parallel, which they are not as they form a triangle OAD):

$$1-t = \frac{2}{5} \implies t = 1 - \frac{2}{5} = \frac{3}{5}$$ $$t = \frac{2}{5}$$

This presents a contradiction, as $t$ cannot be both $\frac{3}{5}$ and $\frac{2}{5}$. This implies that the point F, as defined by the given vector equation, is not on the line AD, despite what the diagram suggests.

This is a common issue in vector problems where the diagram might be illustrative rather than strictly accurate to all given conditions. We must prioritize the algebraic conditions. Therefore, we will proceed with the given $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$ and the collinearity of O, F, E, and E being on AB.

Since O, F, E are collinear, $\vec{OE}$ must be a scalar multiple of $\vec{OF}$. Let $\vec{OE} = k \vec{OF}$ for some scalar $k$.

$$\vec{OE} = k \left( \frac{2}{5}(\vec{a} + \vec{b}) \right) = \frac{2k}{5}\vec{a} + \frac{2k}{5}\vec{b}$$

The diagram shows a line segment from A, through E, to a point that is also labeled E. This is likely a typo, and the point at the end of the line segment from A, through E, should be B. So E is on the line segment AB.
We can write $\vec{OE}$ using the section formula for E on AB:

$$\vec{OE} = (1-s)\vec{OA} + s\vec{OB}$$ $$\vec{OE} = (1-s)\vec{a} + s\vec{OB}$$

We need to determine $\vec{OB}$. The diagram shows D on the line segment OB. This means O, D, B are collinear.
So, $\vec{OB}$ must be a scalar multiple of $\vec{OD}$. Let $\vec{OB} = m \vec{OD} = m\vec{b}$ for some scalar $m$.

$$\vec{OE} = (1-s)\vec{a} + sm\vec{b}$$

$$\frac{2k}{5}\vec{a} + \frac{2k}{5}\vec{b} = (1-s)\vec{a} + sm\vec{b}$$

Equating the coefficients of $\vec{a}$ and $\vec{b}$ (since $\vec{a}$ and $\vec{b}$ are linearly independent):

$$\frac{2k}{5} = 1-s \quad \text{(Equation 1)}$$ $$\frac{2k}{5} = sm \quad \text{(Equation 2)}$$

From Equation 1 and Equation 2, we have:

$$1-s = sm$$ $$1 = s + sm$$ $$1 = s(1+m) \quad \text{(Equation 3)}$$

We have three unknowns ($k, s, m$) and only two independent equations. This indicates that there might be an implicit condition from the diagram that we haven't used, or the problem is designed to be solved by a specific method (e.g., Menelaus' Theorem in vector form).

Let's reconsider the contradiction from Step 1. If the diagram is correct about F being on AD, then the given $\vec{OF}$ is incorrect. If the given $\vec{OF}$ is correct, then F is not on AD. Given the explicit vector equation, we must assume the equation is correct. The diagram's placement of F on AD is therefore misleading.

Let's assume the standard interpretation of such a diagram where O, F, E are collinear, and E is on AB, and D is on OB.
The problem is to express $\vec{DE}$.

$$\vec{DE} = \vec{OE} - \vec{OD} = \vec{OE} - \vec{b}$$

We need to find $\vec{OE}$.

Let's use the property that if O, F, E are collinear, and E is on AB, and D is on OB, then we can use ratios.
From $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$, we know that F is a point.
Since O, F, E are collinear, $\vec{OE} = \lambda \vec{OF} = \lambda \frac{2}{5}(\vec{a} + \vec{b})$.
Since E is on AB, $\vec{OE} = (1-s)\vec{a} + s\vec{OB}$.
Since D is on OB, $\vec{OB} = m\vec{OD} = m\vec{b}$.
So, $\vec{OE} = (1-s)\vec{a} + sm\vec{b}$.
Equating coefficients:
$\frac{2\lambda}{5} = 1-s$
$\frac{2\lambda}{5} = sm$
This implies $1-s = sm \implies 1 = s(1+m)$.

This still leaves us with two unknowns $s$ and $m$.
There must be a missing piece of information or a different interpretation of the diagram.
What if the diagram implies that F is the intersection of OE and AD?
If F is on AD, then $\vec{OF} = (1-t)\vec{a} + t\vec{b}$.
If $\vec{OF} = \frac{2}{5}\vec{a} + \frac{2}{5}\vec{b}$, then $1-t = \frac{2}{5}$ and $t = \frac{2}{5}$. This is a contradiction ($t=\frac{3}{5}$ and $t=\frac{2}{5}$).
This means F, as defined by the equation, cannot be on AD.

Let's assume the diagram is a general representation of a triangle OAB, with D on OB, and E on AB, and F on OE.
The problem asks for $\vec{DE}$.
$\vec{DE} = \vec{OE} - \vec{OD} = \vec{OE} - \vec{b}$.
We have $\vec{OE} = k \vec{OF} = k \frac{2}{5}(\vec{a} + \vec{b})$.
So, $\vec{DE} = k \frac{2}{5}(\vec{a} + \vec{b}) - \vec{b}$.
We need to find $k$.

Consider the line AB. E is on AB.
$\vec{OE} = (1-s)\vec{OA} + s\vec{OB} = (1-s)\vec{a} + s\vec{OB}$.
The diagram shows D on OB. Let $\vec{OB} = m\vec{OD} = m\vec{b}$.
So $\vec{OE} = (1-s)\vec{a} + sm\vec{b}$.
Equating coefficients from $\vec{OE} = \frac{2k}{5}\vec{a} + \frac{2k}{5}\vec{b}$:
$\frac{2k}{5} = 1-s$
$\frac{2k}{5} = sm$
This implies $1-s = sm$.
This is still not enough to find $k$.

Let's re-examine the image for any subtle clues. The point labeled 'E' on the right is the end of the line segment starting from A. This means the line is AB, and E is on AB.
The line segment from O passes through F and then to E.
The line segment from A passes through F and then to D.
This means F is the intersection of OE and AD. This is the standard interpretation.

If F is the intersection of OE and AD, then:

1. F is on AD: $\vec{OF} = (1-t)\vec{OA} + t\vec{OD} = (1-t)\vec{a} + t\vec{b}$.
2. F is on OE: $\vec{OF} = \lambda \vec{OE}$. We are given $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$. This means $(1-t)\vec{a} + t\vec{b} = \frac{2}{5}\vec{a} + \frac{2}{5}\vec{b}$. This implies $1-t = \frac{2}{5}$ and $t = \frac{2}{5}$. This is the contradiction $t = \frac{3}{5}$ and $t = \frac{2}{5}$.

This contradiction means that the problem statement is inconsistent with the diagram if F is the intersection of AD and OE.
Given the explicit vector equation $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$, we must assume this is correct.
The diagram's implication that F is on AD must be incorrect.

Let's assume the problem means:

1. $\vec{OA} = \vec{a}$
2. $\vec{OD} = \vec{b}$
3. $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$
4. E is a point on the line AB (where B is some point such that O, D, B are collinear).
5. O, F, E are collinear.

We need to find $\vec{DE} = \vec{OE} - \vec{OD} = \vec{OE} - \vec{b}$.
Since O, F, E are collinear, $\vec{OE} = k \vec{OF} = k \frac{2}{5}(\vec{a} + \vec{b})$.
Since E is on AB, $\vec{OE} = (1-s)\vec{a} + s\vec{OB}$.
Since O, D, B are collinear, $\vec{OB} = m\vec{OD} = m\vec{b}$.
So $\vec{OE} = (1-s)\vec{a} + sm\vec{b}$.
Equating coefficients:
$\frac{2k}{5} = 1-s$
$\frac{2k}{5} = sm$
This implies $1-s = sm \implies s(1+m) = 1$.

This still leaves us with too many unknowns.
There is a possibility that the question implies that the point D is actually B.
If $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$ (i.e., D is B), then:
$\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$.
E is on AB.
O, F, E are collinear.
If D is B, then we need to express $\vec{BE}$ in terms of $\vec{a}$ and $\vec{b}$.
$\vec{BE} = \vec{OE} - \vec{OB} = \vec{OE} - \vec{b}$.
Since E is on AB, $\vec{OE} = (1-s)\vec{a} + s\vec{b}$.
Since O, F, E are collinear, $\vec{OE} = k \vec{OF} = k \frac{2}{5}(\vec{a} + \vec{b})$.
Equating coefficients:
$1-s = \frac{2k}{5}$
$s = \frac{2k}{5}$
So $1-s = s \implies 1 = 2s \implies s = \frac{1}{2}$.
Then $\frac{2k}{5} = \frac{1}{2} \implies k = \frac{5}{4}$.
So $\vec{OE} = \frac{1}{2}\vec{a} + \frac{1}{2}\vec{b}$.
Then $\vec{DE} = \vec{OE} - \vec{b} = (\frac{1}{2}\vec{a} + \frac{1}{2}\vec{b}) - \vec{b} = \frac{1}{2}\vec{a} - \frac{1}{2}\vec{b} = \frac{1}{2}(\vec{a} - \vec{b})$.
This interpretation makes the problem solvable. It assumes D is B.

Let's verify this interpretation with the diagram. The label 'b' is next to D. If D is B, then $\vec{OB} = \vec{b}$. This is a common convention.
So, let's assume:
$\vec{OA} = \vec{a}$
$\vec{OB} = \vec{b}$ (where B is the point labeled D in the diagram).
F is a point such that $\vec{OF} = \frac{2}{5}(\vec{a} + \vec{b})$.
E is a point on the line segment AB.
O, F, E are collinear.

Since O, F, E are collinear, $\vec{OE} = k \vec{OF}$ for some scalar $k$.

$$\vec{OE} = k \left( \frac{2}{5}(\vec{a} + \vec{b}) \right) = \frac{2k}{5}\vec{a} + \frac{2k}{5}\vec{b}$$

Since E is on the line segment AB, we can write $\vec{OE}$ using the section formula:

$$\vec{OE} = (1-s)\vec{OA} + s\vec{OB}$$ $$\vec{OE} = (1-s)\vec{a} + s\vec{b}$$

$$\frac{2k}{5}\vec{a} + \frac{2k}{5}\vec{b} = (1-s)\vec{a} + s\vec{b}$$

Equating coefficients of $\vec{a}$ and $\vec{b}$:

$$\frac{2k}{5} = 1-s \quad \text{(Equation 1)}$$ $$\frac{2k}{5} = s \quad \text{(Equation 2)}$$

Substitute Equation 2 into Equation 1:

$$s = 1-s$$ $$2s = 1$$ $$s = \frac{1}{2}$$

Now substitute $s = \frac{1}{2}$ back into Equation 2 to find $k$:

$$\frac{2k}{5} = \frac{1}{2}$$ $$2k = \frac{5}{2}$$ $$k = \frac{5}{4}$$

Substitute $k = \frac{5}{4}$ back into the expression for $\vec{OE}$ from Step 1:

$$\vec{OE} = \frac{2}{5} \left( \frac{5}{4} \right) (\vec{a} + \vec{b})$$ $$\vec{OE} = \frac{1}{2}(\vec{a} + \vec{b}) = \frac{1}{2}\vec{a} + \frac{1}{2}\vec{b}$$

We need to find $\vec{DE}$. Remember that we assumed D is the point B, so $\vec{OD} = \vec{b}$.

$$\vec{DE} = \vec{OE} - \vec{OD}$$ $$\vec{DE} = \left( \frac{1}{2}\vec{a} + \frac{1}{2}\vec{b} \right) - \vec{b}$$