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

As we know in saccheri quadrilateral

As you see here $AD=BC,\angle DAB=\angle CBA=90^{\degree}$

$\implies AB=CD,AB||CD\\\implies \angle ADC=\angle BCD=90^{\degree}$

If we join the midpoints of base and summit, then then join

Let M and N be the midpoints of Summit and Base.

In $\Delta AMN$ and $\Delta BMN$

$AN=BN$ ( as N is the midpoint of AB)

$MN=MN$ ( common side)

$AM=BM$ ( digonals of small quad. are equal)

$\therefore \Delta AMN + \Delta BMN$

$\implies \angle BNM=\angle ANM$ ( by C.P.C.T.)

As ANB is straight line

$\therefore \angle BNM+\angle ANM=180^{\degree}\\\implies 2\angle BNM=180^{\degree}\\\implies \angle BNM=90^{\degree}$

So $MN$ is the perpendicular bisector of Base $AB$ as well as og summit $CD.$