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