Given a triangle $ABC$ where $\vec{AC} = \vec{a}$ and $\vec{BC} = \vec{b}$. A square $ACDE$ is constructed on $AC$ and a square $BCFG$ is constructed on $BC$. Given $\vec{AD} = \vec{p}$ and $\vec{BG} = \vec{q}$, express the vectors $\vec{EF}$, $\vec{DF}$, $\vec{FG}$, and $\vec{DE}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{p}$, and $\vec{q}$.

1. Problem Description

Given a triangle $ABC$ where $\vec{AC} = \vec{a}$ and $\vec{BC} = \vec{b}$. A square $ACDE$ is constructed on $AC$ and a square $BCFG$ is constructed on $BC$. Given $\vec{AD} = \vec{p}$ and $\vec{BG} = \vec{q}$, express the vectors $\vec{EF}$, $\vec{DF}$, $\vec{FG}$, and $\vec{DE}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{p}$, and $\vec{q}$.

2. Solution Steps

First, note that since $ACDE$ is a square, we have $\vec{AC} = \vec{DE} = \vec{a}$, $\vec{AD} = \vec{CE} = \vec{p}$, $\vec{AE} = -\vec{CD}$ and $\vec{CD} \perp \vec{AC}$. Similarly, since $BCFG$ is a square, we have $\vec{BC} = \vec{FG} = \vec{b}$, $\vec{BG} = \vec{CF} = \vec{q}$, $\vec{BF} = -\vec{CG}$ and $\vec{CG} \perp \vec{BC}$.

We can express $\vec{EF}$ as

$\vec{EF} = \vec{EC} + \vec{CF} = -\vec{CE} + \vec{CF} = -\vec{p} + \vec{q}$.

We can express $\vec{DF}$ as

$\vec{DF} = \vec{DC} + \vec{CF} = -\vec{CD} + \vec{CF}$.

Since $ACDE$ is a square, $\vec{CD}$ is obtained by rotating $\vec{AC} = \vec{a}$ by $90^{\circ}$ counterclockwise. However, we don't have any information about the rotation. But note that $\vec{AD} = \vec{p}$, therefore we can write $\vec{DC} = \vec{AC} - \vec{AD} = \vec{a} - \vec{p}$. Then,

$\vec{DF} = \vec{DC} + \vec{CF} = \vec{a} - \vec{p} + \vec{q}$.

We are given $\vec{FG} = \vec{b}$, so

$\vec{FG} = \vec{b}$.

We are also given $\vec{DE} = \vec{a}$, so

$\vec{DE} = \vec{a}$.

3. Final Answer

$\vec{EF} = -\vec{p} + \vec{q}$

$\vec{DF} = \vec{a} - \vec{p} + \vec{q}$

$\vec{FG} = \vec{b}$

$\vec{DE} = \vec{a}$