The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

52 NON-EUCLIDEAN GEOMETRY

[CH. III.

Let E, F be the middle points of AB and CD respectively.
Join EF, CE, and DE.
Then the triangles ACE and ED B are congruent, and the congruence of CFE and EFD follows.
Thus the angles at C and D are C D
equal, and EF is perpendicular both
to AB and CD.
Further, the angles at C and D
are acute. f l
To prove this, at C and D draw
C2 and D2 parallel to AB.
Then, by ~ 26 (4), z AC2 =L BDD2. A B
Produce CD to E. FIG. 28.
By ~ 26 (3), ED2 > L DC2.
Therefore, since L ACD =L BDC, it follows that
LEDB> zCDB.
Thus L ACD and L BDC are both acute angles.
~ 29. If in the quadrilateral ABDC, the angles at A and B are
right angles, and the side AC is greater than
BD, the angle at C is less than the angle C
at D. E1 D
Since we are given AC > BD, we can cut
off from AC the segment AE =BD. When
this has been done, join DE.
It follows from ~ 28 that L AED =L BDE. A B
But L ADD > - ACD and L BDC > L BDE. FIG. 29.
Therefore L BDC > z ACD.
The converse of these theorems is easily proved indirectly,
namely, that, if the angles at A and B are right angles, according
as L ACDL BDC, so is ACBD.
~ 30. If ABDC is a quadrilateral in which the angles at A,. B,
and C are right angles, then the angle at D must be acute.
Produce BA through A to B', making AB'= AB. (Fig. 30.)
Draw B'D' perpendicular to B'A and equal to BD.
Join CD', D'A, and DA.
From the congruent triangles D'B'A and DBA, we have
D'A =DA and L D'AB' =L DAB.