isosceles trapezoid

**sanee66** · June 26th 2007, 12:29 PM

Could someone please help me prove that this is an isosceles trapezoid? Thank you in advance.

**Soroban** · June 26th 2007, 02:56 PM

Hello, sanee66!

In quadrilateral MNOA, triangles PNO and PMA are congruent
and APO is an isosceles triangle.
Prove that MNOA is an isosceles triangle.

Code:

                      P
          M * - - - - * - - - - * N
           /     β  *   *  β     \
          /      *        *       \
         /    *              *     \
        /   *                   *   \
       / *  α                   α  * \
    A * - - - - - - - - - - - - - - - * O

Since $\Delta PNO \cong \Delta PMA\!:\;MA = NO$
. . So we must prove that: . $MN \parallel AO$

Since $\Delta PNO \cong \Delta PMA\!:\;\angle MPA = \angle NPO$
. . Let $\angle MPA = \angle NOP = \beta$

We note that: . $2\beta + \angle APO \:=\:180^o$ .[1]

Since $\Delta APO$ is isosceles, $\angle PAO = \angle POA$
. . Let $\angle PAO = \angle POA = \alpha$

In $\Delta APO\!:\;2\alpha + \angle APO \:=\:180^o\quad\Rightarrow\quad \angle APO \:=\:180^o -2\alpha$

Substitute into [1]: . $2\beta + (180^o -2\alpha) \:=\:180^o\quad\Rightarrow\quad \alpha = \beta$

Hence: . $MN \parallel AO$ .(alternate-interior angles)

Therefore, $MNOA$ is an isosceles trapezoid.

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