$A B C D E$ is a pentagon, prove that $A B + B C + C D + D E + E A = 0 .$

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Solution

$\begin{matrix} A p p l y i n g t r i n a g l e r u l e o f v e c t o r a d d i t i o n i n t h e t r i a n g l e A B C, w e g e t - - \to A C = - - \to A B + - - \to B C - - - (1) S i m i l a r l y, i n t r i a n g l e A D E a n d t r i a n g l e A C D w e g e t - - \to D A = - - \to D E + - - \to E A - - - - (2) - - \to A C = - - \to C D + - - \to A D - - - - (3) T h e r e f o r e, - - \to A B + - - \to B C + - - \to C D + - - \to D E + - - \to E A L e t u s g r o u p t h e a b o v e e x p r e s s i o n, (- - \to A B + - - \to B C) + - - \to C D + (- - \to D E + - - \to E A) = - - \to A C + - - \to C D + - - \to D A S u b s t i t u t i n g e q u a t i o n (1) = - - \to A D + - - \to D A = - - \to A D - - - \to A D = 0 H e n c e, p r o v e d . \end{matrix}$

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Similar questions

- - \to A B + - - \to B C + - - \to C D + - - \to D E + - - \to E A = \to 0

\vec{A B} + \vec{B C} + \vec{C D} + \vec{D E} + \vec{E A} = \vec{0}

\vec{A B} + \vec{A E} + \vec{B C} + \vec{D C} + \vec{E D} + \vec{A C} = 3 \vec{A C}

D E = D C, A B = B C = C E = E A = 25 m

41 m

A B C D E

D E = D C, A B = B C = C E = E A = 25 m

41 m

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