Prove a certain quadrilateral is actually a Saccheri Quadrilateral

Question

Question:

"Show that a quadrilateral ABCD, which has angle C = angle D = right angle; and angle A is congruence to angle B, is a Saccheri Quadrilateral."

My attempt:

As by definition, a Saccheri Quadrilateral's (i.e. ABCD) base angles(i.e. Angle C and D) are right angles and also sides AD should be congruence to side BC.

By question, we already have the base angles C and D are equal to right angles. So we just need to show that sides AD and BC are congruence.

I had clue to prove AD // BC by alternate interior angle theorem, but I don't have clue in proving AD congruence to BC.

Please give me some help, thank you!

Answer 1

1. Let $M$ be the midpoint of $CD$
2. Let $l$ be the line perpendicular to $ CD$ at $M$
3. Let $N$ be the intersection point of $l$ and $AB$
4. Since <$NMC$~<$NMD$, $CM$~$DM$, $\Delta$$NMC$~$\Delta$$NMD$ by SAS
5. Hence <$NCM$~<$NDM$
6. Also <$NCB$~<$NDA$ since <$MCB$~<$MDA$ and from 5
7. Since $NC$~$ND$ from 4, by SAA, $\Delta$$BCN$~$\Delta$$ADN$
8. $BC$~$AD$

Here each ~ means "congruent" for segments, angles or triangles.