Final answer:
To demonstrate that EBDF is a Saccheri quadrilateral, we use the properties of ABDC, a Lambert quadrilateral. By extending AB and CD to create AE and CF, and considering that AB≅AE and CD≅CF, we identify that EBDF fulfills the conditions of a Saccheri quadrilateral, having two pairs of equal lengths perpendicular to a common base, with the other pair of opposite sides not necessarily equal.
Step-by-step explanation:
To show that EBDF is a Saccheri quadrilateral, we build on the properties of a Lambert quadrilateral (ABDC), which has three right angles at vertices A, B, and C. By extending sides AB and CD such that AB≅AE and CD≅CF, we create points E and F respectively. The crucial property of a Saccheri quadrilateral is that it has two equal non-congruent sides that are perpendicular to the same base.
In the given figure, since AB≅AE and AD is perpendicular to both AB and AE by the definition of a Lambert quadrilateral and its extended form, DE is a straight line which implies that DAE is a straight angle. Similarly, since CD≅CF and BC is perpendicular to both CD and CF, BF is a straight line, creating a straight angle at BCF. Thus, we observe that EBDF has two pairs of equal lengths (AE = AB and CF = CD) and each pair is perpendicular to a common base (AB or CD extended).
Therefore, EBDF is indeed a Saccheri quadrilateral as it has all the defining traits: two equal opposite sides (EB = DF), both perpendicular to the same base (BF and DE are straight lines), and the other pair of opposite sides (ED and BF) are not necessarily congruent.