ABCD is a quadrilateral. Show that $- - \to A B + - - \to A D + - - \to C B + - - \to C D = 4 - - \to P Q$ $vec(AB)+vec(AD)+vec(CB)+vec(CD)=4vec(PQ)$ where P and Q are the mid points of AC and BD respectively?

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ABCD is a quadrilateral. Show that $- - \to A B + - - \to A D + - - \to C B + - - \to C D = 4 - - \to P Q$ $vec(AB)+vec(AD)+vec(CB)+vec(CD)=4vec(PQ)$ where P and Q are the mid points of AC and BD respectively?

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Answer 1 · 2018-03-14T21:40:15

That's true

Explanation:

https://www.geogebra.org/m/WgaNJ5Xq

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$- - \to A B + - - \to A D = 2 \cdot - - \to A Q$ $\vec (AB)+\vec (AD)= 2 \cdot \vec (AQ)$
$- - \to A B + - - \to B C = 2 \cdot - - \to A P$ $\vec (AB)+\vec (BC)= 2 \cdot \vec (AP)$
$- - \to P Q = - - \to A Q - - - \to A P$ $\vec (PQ)=\vec (AQ)- \vec (AP)$

$2 \cdot - - \to P Q = - - \to A D + - - \to C B$ $2 cdot \vec(PQ)=\vec (AD)+\vec(CB)$
$2 \cdot - - \to P Q = - - \to A B + - - \to C D$ $2 cdot \vec(PQ)=\vec (AB)+\vec(CD)$

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ABCD is a quadrilateral. Show that $- - \to A B + - - \to A D + - - \to C B + - - \to C D = 4 - - \to P Q$ $vec(AB)+vec(AD)+vec(CB)+vec(CD)=4vec(PQ)$ where P and Q are the mid points of AC and BD respectively?

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ABCD is a quadrilateral. Show that −−→AB+−−→AD+−−→CB+−−→CD=4−−→PQvec(AB)+vec(AD)+vec(CB)+vec(CD)=4vec(PQ) where P and Q are the mid points of AC and BD respectively?

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ABCD is a quadrilateral. Show that $- - \to A B + - - \to A D + - - \to C B + - - \to C D = 4 - - \to P Q$ $vec(AB)+vec(AD)+vec(CB)+vec(CD)=4vec(PQ)$ where P and Q are the mid points of AC and BD respectively?