Prove that AB=AC given that ABD= ACE.

Given that P is a point on OA such that $OP = \frac{1}{3} OA$. Since $\vec{OA} = \mathbf{a}$, we have:

Given that Q is a point on OB such that $OQ = \frac{1}{2} OB$. Since $\vec{OB} = \mathbf{b}$, we have:

Given that R is a point on AB such that $AR = \frac{1}{3} AB$.
First, find the vector $\vec{AB}$:

Now, express $\vec{AR}$:

Using vector addition, $\vec{OR} = \vec{OA} + \vec{AR}$:

Using the relation $\vec{PR} = \vec{OR} - \vec{OP}$:

Using the relation $\vec{RQ} = \vec{OQ} - \vec{OR}$:

From Step 4, $\vec{PR} = \frac{1}{3} \mathbf{a} + \frac{1}{3} \mathbf{b}$.
From Step 5, $\vec{RQ} = -\frac{2}{3} \mathbf{a} + \frac{1}{6} \mathbf{b}$.
Since the coefficients of $\mathbf{a}$ and $\mathbf{b}$ are not equal in both expressions, $\vec{PR} \neq \vec{RQ}$.
Therefore, the statement $\vec{PR} = \vec{RQ}$ cannot be shown to be true under the given conditions.