Let ABCD be a parallelogram whose diagonals intersect at P and ley O be the origin, then OA^(→) +OB^(→) +OC^(→) +OD^(→) equals


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Question Number 41291 by Mayur last updated on 04/Aug/18

$Let A B C D be a parallelogram whose$ $diagonals intersect at P and ley O be$ $the origin, then \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D}$ $equals$
	Answered by MJS last updated on 04/Aug/18

	$4 \overset{⇀}{O P}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

	$O \vec{A} + A \vec{P} = O \vec{P}$ $O \vec{B} + B \vec{P} = O \vec{P}$ $O \vec{C} - P \vec{C} = O \vec{P}$ $O \vec{D} - P \vec{D} = O \vec{P}$ $n o w a d d t h e m$ $O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} + (A \vec{P} - P \vec{C}) + (B \vec{P} - P \vec{D}) = 4 O \vec{P}$ $O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} = 4 O \vec{P}$ $1 . (s i n c e O \vec{P} + P \vec{D} = O D)$ $2 . ∣ A \vec{P} ∣=∣ P \vec{C} ∣$ $3 . ∣ \vec{B P} ∣=∣ P \vec{D} ∣ d i o a g o n a l b i s e c t e a c h o t h e r$
	Commented by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

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