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

3] PROBLEMS OF CONSTRUCTION 7
'PROBLEM 3. At a given point on a given straight line to erect the
perpendicular.
Construction. Let A be the given point and BC the given straight line.
Draw the perpendicular ZOY (by Problem 2), meeting BC in O.
Take OY = OZ, and join AY and AZ.
Produce YA through A to X.
Bisect LXAZ by AD (by Problem 1).
Then AD is the perpendicular to BC D
through A. /
Proof. By the construction, the tri-
angles OAZ and OAY are congruent.
Therefore L ZAO = L YAO B 0 A C
= L.XAC.
But L DAZ= L XAD. F. 3.
Therefore AD is perpendicular to BC.
PROBLEM 4. From a given point outside a given straight line to
draw the perpendicular to the line.
Construction. Let A, B be two points on the given line, and C the
point outside it.
Join AC and BC.
On the segment AB take a point D, and
(by Problem 3) draw the perpendicular at
D to AB.
By Pasch's Axiom, this line must cut
either AC or BC. E
Let it cut AC, and let the point of
intersection be E.
Produce ED through D to F, so that
DE=DF. H B
Join AF and produce AF to G, such
that AG =AC.
Join CG, and let it be cut by AB, or AB F
produced, at H.
Then CH is the required perpendicular.
Proof. From the construction, the triangles ADE and ADF are congruent, so FIG. 4.
that AB bisects LCAG.
It follows that the triangles ACH and AGH are congruent, and that
LAHC is a right angle.
PROBLEM 5. At a given point on a given straight line to make an
angle equal to a given angle.
Construction. Let A be the point on the given line a. (Cf. Fig. 5.)
Let D be the given (acute) angle.
From a point E on one of the lines bounding the angle, draw (by
Problem 4) the perpendicular EF to the other bounding line.