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 $\vec{q}$.

GeometryVectorsGeometry of SquaresVector AdditionGeometric Proof
2025/3/27

1. Problem Description

Given a triangle ABCABC where AC=a\vec{AC} = \vec{a} and BC=b\vec{BC} = \vec{b}. A square ACDEACDE is constructed on ACAC and a square BCFGBCFG is constructed on BCBC. Given AD=p\vec{AD} = \vec{p} and BG=q\vec{BG} = \vec{q}, express the vectors EF\vec{EF}, DF\vec{DF}, FG\vec{FG}, and DE\vec{DE} in terms of a\vec{a}, b\vec{b}, p\vec{p}, and q\vec{q}.

2. Solution Steps

First, note that since ACDEACDE is a square, we have AC=DE=a\vec{AC} = \vec{DE} = \vec{a}, AD=CE=p\vec{AD} = \vec{CE} = \vec{p}, AE=CD\vec{AE} = -\vec{CD} and CDAC\vec{CD} \perp \vec{AC}. Similarly, since BCFGBCFG is a square, we have BC=FG=b\vec{BC} = \vec{FG} = \vec{b}, BG=CF=q\vec{BG} = \vec{CF} = \vec{q}, BF=CG\vec{BF} = -\vec{CG} and CGBC\vec{CG} \perp \vec{BC}.
We can express EF\vec{EF} as
EF=EC+CF=CE+CF=p+q\vec{EF} = \vec{EC} + \vec{CF} = -\vec{CE} + \vec{CF} = -\vec{p} + \vec{q}.
We can express DF\vec{DF} as
DF=DC+CF=CD+CF\vec{DF} = \vec{DC} + \vec{CF} = -\vec{CD} + \vec{CF} .
Since ACDEACDE is a square, CD\vec{CD} is obtained by rotating AC=a\vec{AC} = \vec{a} by 9090^{\circ} counterclockwise. However, we don't have any information about the rotation. But note that AD=p\vec{AD} = \vec{p}, therefore we can write DC=ACAD=ap\vec{DC} = \vec{AC} - \vec{AD} = \vec{a} - \vec{p}. Then,
DF=DC+CF=ap+q\vec{DF} = \vec{DC} + \vec{CF} = \vec{a} - \vec{p} + \vec{q}.
We are given FG=b\vec{FG} = \vec{b}, so
FG=b\vec{FG} = \vec{b}.
We are also given DE=a\vec{DE} = \vec{a}, so
DE=a\vec{DE} = \vec{a}.

3. Final Answer

EF=p+q\vec{EF} = -\vec{p} + \vec{q}
DF=ap+q\vec{DF} = \vec{a} - \vec{p} + \vec{q}
FG=b\vec{FG} = \vec{b}
DE=a\vec{DE} = \vec{a}

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