help proving midpoints of quadrilaterals

**johng** · February 18th 2013, 03:23 PM

For both questions, look for congruent triangles. Once you've identified the congruent triangles, study the angles.

**djo4567** · February 18th 2013, 05:06 PM

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**Soroban** · February 18th 2013, 05:44 PM

Hello, djo4567!

I'll do the first one.

Connect the midpoint of the sides of a rhombus.
Prove that the midpoints form a rectangle.

$\text{We have rhombus }ABCD,\,\text{ and midpoints }E, F, G, H.$

Code:

                      E
            A o * * * o * * * o B
             *     *   *     *
            *   *       *   *
           * *           * *
        H o               o F
         * *           * *
        *   *       *   *
       *     *   *     *
    D o * * * o * * * o C
              G

$\text{Draw diagonal }BD.$

$\text{In }\Delta ABD,\,E\text{ and }H\text{ are midoints of }AB\text{ and }AD.}$
$\text{Hence: }\,EH = \tfrac{1}{2}BD\text{ and }EH \parallel BD.$
$\text{(Theorem: the segment joining the midpoints of two sides of a triangle}$
. . $\text{is parallel to and one-half the length of the third side.)}$

$\text{In }\Delta CBD,\,F\text{ and }G\text{ are midpoints of }CB\text{ and }CD.$
$\text{Hence: }\,FG = \tfrac{1}{2}BD\text{ and }FG \parallel BD.$

$\text{Hence, }EFGH\text{ is a parallelogram.}$
$\text{(Theorem: if two sides of a quadrilateral are equal and parallel,}$
. . $\text{the quadrilateral is a parallelogram.)}$

$\text{Draw diagonal }AC.$
$\text{Then: }\,AC \perp BD.$
$\text{(Diagonals of a rhombus are perpendicular..)}$

$\text{In }\Delta ABC,\,EF = \tfrac{1}{2} AC\text{ and }EF \parallel AC.$

$\text{Since }EH \parallel BD,\,EF \perp EH.$

$\text{Therefore, }EFGH\text{ is a rectangle.}$

**djo4567** · February 18th 2013, 07:09 PM

Originally Posted by Soroban

Hello, djo4567!

I'll do the first one.

$\text{We have rhombus }ABCD,\,\text{ and midpoints }E, F, G, H.$

Code:

                      E
            A o * * * o * * * o B
             *     *   *     *
            *   *       *   *
           * *           * *
        H o               o F
         * *           * *
        *   *       *   *
       *     *   *     *
    D o * * * o * * * o C
              G

$\text{Draw diagonal }BD.$

$\text{In }\Delta ABD,\,E\text{ and }H\text{ are midoints of }AB\text{ and }AD.}$
$\text{Hence: }\,EH = \tfrac{1}{2}BD\text{ and }EH \parallel BD.$
$\text{(Theorem: the segment joining the midpoints of two sides of a triangle}$
. . $\text{is parallel to and one-half the length of the third side.)}$

$\text{In }\Delta CBD,\,F\text{ and }G\text{ are midpoints of }CB\text{ and }CD.$
$\text{Hence: }\,FG = \tfrac{1}{2}BD\text{ and }FG \parallel BD.$

$\text{Hence, }EFGH\text{ is a parallelogram.}$
$\text{(Theorem: if two sides of a quadrilateral are equal and parallel,}$
. . $\text{the quadrilateral is a parallelogram.)}$

$\text{Draw diagonal }AC.$
$\text{Then: }\,AC \perp BD.$
$\text{(Diagonals of a rhombus are perpendicular..)}$

$\text{In }\Delta ABC,\,EF = \tfrac{1}{2} AC\text{ and }EF \parallel AC.$

$\text{Since }EH \parallel BD,\,EF \perp EH.$

$\text{Therefore, }EFGH\text{ is a rectangle.}$

THANK YOU!!!! That helped soo much.

**johng** · February 19th 2013, 07:48 AM

Here's a second proof obtained just by considering angles; I think it's more elementary than the above solution:

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