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