Let ABCD be a quadrilateral. Suppose AB and CD have equal length and angles BAD and BCD are supplementary (i.e., angle BAD plus angle BCD equals 180 degrees). Show that AD is parallel to BC.

Construct diagonals AC and BD intersecting at point E.

  <BAD + <BCD = 180 ==> ABCD is cyclic ==>

    1) <EAB = <CAB = <CDB = <CDE = <EDC
    2) <EBA = <DBA = <DCA = <DCE = <ECD
    3) <DAE = <DAC = <DBC

1) and 2) with AB = CD ==> triangles EAB and EDC
are congruent ==> EA = ED ==> <ADE = <DAE

Therefore, combining this with 3) gives

<ADB = <ADE = <DAE = <DAC = <DBC

Thus, AD and BC are parallel.

Posted by Bractals on 2005-05-29 01:50:35