Given a triangle $ABC$ with $\vec{AC} = \vec{a}$ and $\vec{BC} = \vec{b}$. A square $ACDE$ is constructed on $AC$ and a square $BCFG$ is constructed on $BC$. If $\vec{AD} = \vec{p}$ and $\vec{BG} = \vec{q}$, determine 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 $ABC$ with $\vec{AC} = \vec{a}$ and $\vec{BC} = \vec{b}$. A square $ACDE$ is constructed on $AC$ and a square $BCFG$ is constructed on $BC$. If $\vec{AD} = \vec{p}$ and $\vec{BG} = \vec{q}$, determine 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

Since $ACDE$ is a square, we have $AC = CD = DE = EA$ and $AC \perp CD$, $CD \perp DE$, $DE \perp EA$, $EA \perp AC$. Similarly, since $BCFG$ is a square, we have $BC = CF = FG = GB$ and $BC \perp CF$, $CF \perp FG$, $FG \perp GB$, $GB \perp BC$.

Also, $\vec{AC} = \vec{a}$, $\vec{AD} = \vec{p}$, $\vec{BC} = \vec{b}$, $\vec{BG} = \vec{q}$.

Since $ACDE$ is a square, $\vec{AC} \perp \vec{AD}$ and $|AC|=|AD|$, $\vec{CD} = R_{\pi/2} \vec{AC}$. But we don't know if the rotation is clockwise or counter-clockwise. So, $\vec{CD} = \pm R_{\pi/2} \vec{a}$.

Similarly, $\vec{CF} = \pm R_{\pi/2} \vec{BC} = \pm R_{\pi/2} \vec{b}$. We assume here that the squares are constructed outside of the triangle.

Since $ACDE$ is a square, $\vec{AE} = -\vec{CD}$. So $\vec{DE} = -\vec{AC} = -\vec{a}$.

Also, $\vec{DE} = -\vec{a}$.

$\vec{EF} = \vec{EA} + \vec{AC} + \vec{CF} = -\vec{AD} + \vec{AC} + \vec{CF} = -\vec{p} + \vec{a} + \vec{CF}$. Since BCFG is a square, $\vec{CF} = \pm R_{\pi/2} \vec{BC} = \pm R_{\pi/2} \vec{b}$. Let's assume we have $\vec{CF} = R_{\pi/2} \vec{b}$. Then $\vec{EF} = \vec{a} - \vec{p} + R_{\pi/2} \vec{b}$. Also, since $\vec{AD} \perp \vec{AC}$ and $|AD|=|AC|$ then $p \perp a$, but this is not a vector relationship.

Since $\vec{AD} = \vec{p}$, $\vec{AC} = \vec{a}$ and $ACDE$ is a square, $\vec{DE} = -\vec{a}$ and $\vec{AE} = -\vec{CD}$. Since $BCFG$ is a square, $\vec{FG} = -\vec{b}$ and $\vec{CF}$ is orthogonal to $\vec{b}$ and $|\vec{CF}| = |\vec{b}|$. Since $\vec{ACDE}$ is a square, $\vec{CD}$ is orthogonal to $\vec{a}$ and $|\vec{CD}|=|\vec{a}|$. Let $\vec{CD} = \vec{a}^{\perp}$, then $\vec{AD} = \vec{p} = \vec{a}^{\perp}$. Similarly, $\vec{CF} = \vec{b}^{\perp}$ where $|\vec{b}^{\perp}| = |\vec{b}|$ and $\vec{b}^{\perp} \perp \vec{b}$.

$\vec{EF} = \vec{EA} + \vec{AC} + \vec{CF} = -\vec{CD} + \vec{AC} + \vec{CF} = - \vec{a}^{\perp} + \vec{a} + \vec{b}^{\perp}$.

$\vec{DF} = \vec{DC} + \vec{CB} + \vec{BF} = \vec{AD}$ where $\vec{AD} = \vec{p}$, $\vec{DC} = -\vec{CD}= -a^{\perp}$, $\vec{BF} = -\vec{FG} = \vec{b}$, $\vec{CB} = -b$. Then $\vec{DF} = -\vec{a}^{\perp} - \vec{b} + \vec{q}$. $\vec{DF} = DC + CB + BF$ = $- CD - BC + BG$. Since $CD \perp a$, $CD$ rotates $a$ by $90 degrees$, $BG \perp b$ and $BG$ rotates $b$ by $90 degrees$.

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

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

3. Final Answer

$\vec{EF} = \vec{a} - \vec{p} + \vec{b}^\perp$

$\vec{DF} = -\vec{a}^\perp - \vec{b} + \vec{q}$.

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

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