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

6 NON-EUCLIDEAN GEOMETRY [CH. i.
PROBLEM 1. To bisect a given angle.
Construction. On one of the lines bounding the given angle A take
any two points B, C.
On the other bounding line take AB'= AB and AC' AC.
Join BC' and B'C.
Let them intersect at D.
Then AD is the desired bisector.
C
A C
FIG. 1.
Proof. The triangles BAC' and B'AC are congruent.
Therefore L/ACB'=/ AC'B and L DBC= LDB'C'.
It follows that the triangles BDC and B'DC' are congruent, since
BC= B'C'.
Therefore DB'=DB.
Finally the triangles BAD and B'AD are congruent, and AD bisects
the given angle.
PROBLEM 2. To draw a perpendicular to a given straight line.
Construction. Let AB be the given straight line.
Take any other straight line AC through
A.
Upon AB take AD=AC. H
Join CD.
Bisect LCAD (by Problem 1), and let the C G
bisector cut CD at G.
On AB take AF =AG, and on the ray AG
take AH=AD. A F D B
Join FH. FIG. 2.
Then FH is perpendicular to AB.
Proof. From the triangles ACG and ADG, we have L AGD equal to
a right angle.
Also the triangles AGD and AFH are congruent.
Therefore L AFH = LAGD= 1 right angle.