Question

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 .

Answer 1

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