Vector geom.: If the midpoints of the consecutive sides ...

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If the midpoints of the consecutive sides of any parallelogram are connected by striaght lines, prove by using vectors that the resulting quadrilateral is a parallelogram.

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Re: Vector geometry question

Hello, americo74!

If the midpoints of the consecutive sides of any parallelogram are connected by striaght lines,
prove by using vectors that the resulting quadrilateral is a parallelogram.
Code:
                   A       P           B
                   *- - - -*- - - - - -*
                  /     *    *        /
                 /   *         *     /
                / *              *  /
             S *                   * Q
              /  *              * /
             /     *         *   /
            /        *    *     /
         D-*- - - - - -*- - - -* C
                       R

We have parallelogram ABCD\displaystyle ABCD with midpoints P,Q,R,S.\displaystyle P,\,Q,\,R,\,S.
Draw diagonal DB.\displaystyle DB.

We have: SP=SA+AP=12DA+12AB=12(DA+AB)=12DB\displaystyle \,\vec{SP} \:=\:\vec{SA}\,+\,\vec{AP} \:=\:\frac{1}{2}\vec{DA}\,+\,\frac{1}{2}\vec{AB}\:=\:\frac{1}{2}\left(\vec{DA}\,+\,\vec{AB}\right) \:=\:\frac{1}{2}\vec{DB}
. . Hence: SPDB\displaystyle SP\,\parallel\,DB\, and SP=12DB\displaystyle \,|SP|\,=\,\frac{1}{2}|DB|

We have: RQ=RC+CQ=12DC+12CB=12(DC+CB)=12DB\displaystyle \,\vec{RQ}\:=\:\vec{RC}\,+\,\vec{CQ}\:=\:\frac{1}{2}\vec{DC}\,+\,\frac{1}{2}\vec{CB}\:=\:\frac{1}{2}\left(\vec{DC}\,+\,\vec{CB}\right)\:=\:\frac{1}{2}\vec{DB}
. . Hence: RQDB\displaystyle \,RQ\,\parallel\,DB\, and RQ=12DB\displaystyle \,|RQ| \,= \,\frac{1}{2}|DB|

Then we have: SPRQ\displaystyle \,SP\,\parallel\,RQ\, and SP=RQ\displaystyle \,|SP|\,=\,|RQ|

Theorem: if two sides of a quadrilateral are parallel and equal,
. . . . . . . .the quadrilateral is a parallelogram.

Therefore, PQRS\displaystyle PQRS is a parallelogram.

 
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