A) Express the following vectors in terms of b and c only. i) AE ii) BD b) By expressing vectors BF in two ways, find the ratio BF:FD given that BF = hBD and AF = Tae c) Hence find vector BF in terms of b and c only

Since E is the midpoint of BC,

$\boxed{\text{i) } \vec{AE} = \frac{1}{2}\mathbf{b} + \frac{1}{2}\mathbf{c}}$

Since D is the midpoint of AC,

$\boxed{\text{ii) } \vec{BD} = -\mathbf{b} + \frac{1}{2}\mathbf{c}}$

We are given $\vec{AF} = t\vec{AE}$. Substitute $\vec{AE}$ from part (a) i.

We use the path $\vec{AF} = \vec{AB} + \vec{BF}$. We are given $\vec{BF} = h\vec{BD}$.
Substitute $\vec{BD}$ from part (a) ii.

Now substitute this into the expression for $\vec{AF}$:

Since $\mathbf{b}$ and $\mathbf{c}$ are non-parallel vectors, their coefficients must be equal.

Comparing coefficients of $\mathbf{b}$:

Comparing coefficients of $\mathbf{c}$:

From Equation 4, we get $t = h$.
Substitute $t=h$ into Equation 3:

Multiply both sides by 2:

Since $\vec{BF} = h\vec{BD}$, this means $\vec{BF} = \frac{2}{3}\vec{BD}$.
The ratio BF:FD is $h : (1-h)$.

$\boxed{\text{BF : FD} = 2 : 1}$

We found $h = \frac{2}{3}$.
We know $\vec{BF} = h\vec{BD}$.

$\boxed{\vec{BF} = -\frac{2}{3}\mathbf{b} + \frac{1}{3}\mathbf{c}}$