Question: In the given diagram ABCD is a pentagon, prove that: \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA} = 0
Solution:
To prove: \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA} = 0

Using ∆ law of vector addition in ∆ ABC,
\vec{AC} = \vec{AB} + \vec{BC} - (i)
Using ∆ law of vector addition in ∆ ADE,
\vec{DA} = \vec{DE} + \vec{EA} - (ii)
Using ∆ law of vector addition in ∆ ADC,
\vec{DC} = \vec{DA} + \vec{AC} - (iii)
Taking LHS,
= \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA}
[ Replace \vec{AC} = \vec{AB} + \vec{BC} from equation (i) ]
= \vec{AC} + \vec{CD} + \vec{DE} + \vec{EA}
[ Replace \vec{DA} = \vec{DE} + \vec{EA} from equation (ii) ]
= \vec{AC} + \vec{CD} + \vec{DA}
= \vec{AC} + \vec{DA} + \vec{CD}
[ Replace \vec{DC} = \vec{DA} + \vec{AC} from equation (iii) ]
= \vec{DC} + \vec{CD}
= \vec{DC} - \vec{DC}
= 0
= RHS
#proved
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