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Angle between Pair of Straight Lines

A B C D E is ...

Question

ABCDE is a pentagon, prove that
(i) $\vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A} = \vec{0}$

(ii) $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C} = 3 \vec{A C}$

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Solution

Given: $A B C D E$ is a pentagon.
(i) To Prove: $\vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A} = \vec{0} .$
Proof: We have,
$LHS = \vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A}$
$= \vec{A C} + \vec{C D} + \vec{D A}$ [ ∵ $\vec{A B} + \vec{B C} = \vec{A C}$ and $\vec{D E} + \vec{E A} = \vec{A D}$ ]
$= \vec{A D} + \vec{D A}$ [ ∵ $\vec{A C} + \vec{C D} = \vec{A D}$ ]
$= \vec{0}$ = RHS
(ii) To Prove: $\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C} = 3 \vec{A C} .$
Proof: We have,
$LHS = \vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C}$
$= (\vec{A B} + \vec{B C}) + (\vec{A E} + \vec{E D}) + \vec{D C} + \vec{A C}$
$= \vec{A C} + \vec{A D} + \vec{D C} + \vec{A C}$ [∵ $\vec{A B} + \vec{B C} = \vec{A C}$ and $\vec{A E} + \vec{E D} = \vec{A D}$ ]
= $\vec{A C} + \vec{A D} + \vec{A C} - \vec{A D} + \vec{A C}$ [∵ $\vec{A D} + \vec{D C} = \vec{A C} \Rightarrow \vec{D C} = \vec{A C} - \vec{A D}$ ]
$= 3 \vec{A C} = RHS$
Hence proved.

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