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 Disguised trapezoid (Posted on 2005-05-28)
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.

 See The Solution Submitted by McWorter Rating: 4.2000 (5 votes)

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 Cyclic Comment 3 of 3 |
` `
`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 EDCare 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

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