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

8

NON-EUCLIDEAN GEOMETRY

[CH. l.

On Aa take AC= DF.
At C erect the perpendicular Cc to Aa (by Problem 3).
Make BC=EF, and join AB.
E
X/ / ysB
D F A Ca
FIG. 5.
Proof. By the construction, the triangles DEF and ABC are
congruent.
Therefore L BAC= L EDF.
PROBLEM 6. To bisect a given finite straight line.
Construction. Let AB be the given segment. b
At B draw the perpendicular Bb to AB (by
Problem 3). C
Upon Bb take any point C and join AC,
At B make L ABE = L BAC (by Problem 5).
Let the line BE cut AC at D.
Bisect LADB by the line cutting AB at F (by
Problem 1).
Then F is the middle point of AB.
Proof. From the construction it follows that A F B
the triangles ADF and DBF are congruent. FIG. 6.
Thus AF=FB.
Note. This construction has to be slightly modified for the Elliptic
Geometry. The point C must lie between B and the pole of AB.
[Cf. ~ 78.]
~ 4. Two Theorems independent of the Parallel
Postulate.
1. The perpendicular to the base of any triangle through its
middle point is also perpendicular to the line joining the middle
points of the two sides.
Let ABC be any triangle, and let F and E be the middle
points of the sides AB and AC.
Join F and E; and draw AA', BB', and CC' perpendicular
to EF from A, B, and C.
Let H be the middle point of BC, and K the middle point
of B'C'.
Join HK.