ABCDE is a pentagon, prove that, AB + BC + CD + DE + EA = 0 - Vidyadip

Question

ABCDE is a pentagon, prove that,

$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$

✓

Answer

Given ABCDE is a pentagon.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\Big(\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}\Big)+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$
$=\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ABC},\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\Big]$
$=\Big(\overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}\Big)+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}$ $\Big[$Using triangle law in $\triangle\text{ACD},\ \overrightarrow{\text{AC}}+\overrightarrow{\text{CD}}=\overrightarrow{\text{AD}}\Big]$
$=\overrightarrow{\text{AD}}+\overrightarrow{\text{DA}}$
$=\overrightarrow{\text{AD}}-\Big(-\overrightarrow{\text{AD}}\Big)$
$=\vec0$
$\therefore\ \overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CD}}+\overrightarrow{\text{DE}}+\overrightarrow{\text{EA}}=\vec0$

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