Answer to Question #143186 in Geometry for J

Question #143186

prove that the line joining the midpoints of the base and summit of a saccheri quadrilateral is the perpendicular bisector of both the base and the summit .


1
Expert's answer
2020-11-12T18:23:25-0500


Proof. Let E and F be the midpoints of the base and of a saccheri quadrilateral respectively. Note \triangleDEF \cong \triangle CEB by SAS. Thus DE\overline{DE } \cong CE\overline{CE } , \angle AED=\angle CED and \angle ADE=\angle BCE. Invoking SSS, it follows that \triangle DEF \cong\triangle CEF. Hence \angle DFE=\angle CFE, and since these angles are supplementary, each must be 90. Also, \angle DEF=\angle CEF, and so \angle AED+\angle DEF =\angle CED+\angle DEF=90, again since these angles are supplementary. Thus EF\overline{EF }AB\overline{AB } and EF\overline{EF }CD\overline{CD} , and it follows from Proposition 27 that AB\overleftrightarrow{AB} \|CD\overleftrightarrow{CD} with common perpendicular EF\overleftrightarrow{EF} .


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Comments

Assignment Expert
12.06.21, 14:58

Dear Anne, please use the panel for submitting a new question.


Anne
24.05.21, 06:54

What about on how to prove the summit and base of Saccheri quadrilateral are paralle?

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