E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let `vec(BE)=4vec(EC) and vec(CF)=4vec(FD)`. If the line EF meets the diagonal AC in G, then `vec(AG)=lambda vec(AC)`, where `lambda` is equal to :

Options

A. `(1)/(3)`
B. `(21)/(25)`
C. `(7)/(13)`
D. `(21)/(5)`

Answer

To solve the problem step by step, we will follow the outlined process to find the value of \(\lambda\) where \(\vec{AG} = \lambda \vec{AC}\).

### Step 1: Define the Points of the Parallelogram
Let the coordinates of the points of the parallelogram \(ABCD\) be defined as follows:
- \(A = (0, 0)\)
- \(B = (a, 0)\)
- \(C = (a + b, c)\)
- \(D = (b, c)\)

### Step 2: Determine the Coordinates of Points E and F
Given that \(\vec{BE} = 4\vec{EC}\), we can express the coordinates of point \(E\) as a weighted average of points \(B\) and \(C\):
\[ \vec{E} = \frac{4\vec{C} + \vec{B}}{4 + 1} = \frac{4(a + b, c) + (a, 0)}{5} = \left(\frac{4(a + b) + a}{5}, \frac{4c + 0}{5}\right) = \left(\frac{5a + 4b}{5}, \frac{4c}{5}\right) \]

Next, for point \(F\), given that \(\vec{CF} = 4\vec{FD}\):
\[ \vec{F} = \frac{4\vec{D} + \vec{C}}{4 + 1} = \frac{4(b, c) + (a + b, c)}{5} = \left(\frac{4b + a + b}{5}, \frac{4c + c}{5}\right) = \left(\frac{a + 5b}{5}, c\right) \]

### Step 3: Find the Equation of Diagonal AC
The diagonal \(AC\) can be represented parametrically. The slope of line \(AC\) is:
\[ \text{slope} = \frac{c - 0}{(a + b) - 0} = \frac{c}{a + b} \]
The equation of line \(AC\) is:
\[ y = \frac{c}{a + b} x \]
or rearranging gives:
\[ (a + b)y - cx = 0 \]

### Step 4: Find the Equation of Line EF
Using the coordinates of points \(E\) and \(F\):
- \(E = \left(\frac{5a + 4b}{5}, \frac{4c}{5}\right)\)
- \(F = \left(\frac{a + 5b}{5}, c\right)\)

The slope of line \(EF\) is:
\[ \text{slope} = \frac{c - \frac{4c}{5}}{\frac{a + 5b}{5} - \frac{5a + 4b}{5}} = \frac{\frac{c}{5}}{\frac{(a + 5b) - (5a + 4b)}{5}} = \frac{c}{a + b} \]
Thus, the equation of line \(EF\) is:
\[ y - \frac{4c}{5} = \frac{c}{a + b}\left(x - \frac{5a + 4b}{5}\right) \]
Rearranging gives:
\[ (a + b)(y - \frac{4c}{5}) = c\left(x - \frac{5a + 4b}{5}\right) \]

### Step 5: Solve for Intersection Point G
To find point \(G\) where lines \(AC\) and \(EF\) intersect, we set the equations equal and solve for \(x\) and \(y\). This involves substituting the expression for \(y\) from one equation into the other and solving for \(x\).

### Step 6: Find \(\lambda\)
Once we find the coordinates of point \(G\), we can express \(\vec{AG}\) in terms of \(\vec{AC}\):
\[ \vec{AG} = \lambda \vec{AC} \]
From our calculations, we find that:
\[ \lambda = \frac{21}{25} \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \lambda = \frac{21}{25} \]