Particular Enunciation.

Given.—The straight line AB and the point C in AB

Required.–To draw from the point ca straight line at right angles to A B. Construction.

(a) In A C take any point D. (6) From CB cut off a part CE equal to CD.

(Euc. I. 3.)

(c) On D E describe the equilateral triangle

DFE. (Euc. I. 1.) (d) Join FC. The straight line FC drawn from the given

point C, shall be at right angles to the given straight line A B.

If FC be at right angles to A B, we must prove

that angle FCD is equal to angle FCE. Proof (with Syllogisms in full). D C equals CE (by Construction 6);

add to each CF. (e) Then by Axiom 2a, DC, CF are equal

to EC, CF, each to each. Ist Syllogism. If two triangles have two sides of the one

equal to two sides of the other, and the base of the one equal to the base of the other; those sides shall contain equal angles

(Euc. I. 8). The two triangles FCD, FCE have the two

sides D C, C F equal to the two sides EC, CF (e), each to each, and they have the base D equal to the base E F (by construction (c),

FDC being an equilateral triangle). (f).. the angle DCF is equal to the angle

ECF. 2nd Syllogism. When a straight line, standing on another straight

line, makes the adjacent angles equal, each of the angles is called a right angle

(Euc. Def. 10). The straight line F C standing on the straight line

AB makes the angle DCF equal to the

adjacent angle ECE (f). .. Each of the angles DCF, ÉCF is called

a right angle. Result.-Wherefore from the given point C, a line CF has been drawn at right angles to A B.

C.E.F. EXERCISES.—I. Write out this proof in contracted syllogisms.

II. At the point N in the given

N straight line N O draw a straight line at right angles to NO (produce N O towards N).

III. At the point N in the straight line given in (II), draw a straight line double the length of N O at right angles to No. (Produce N O both ways; make the produced parts each equal to No; draw a line at right angles from N; describe a circle with N as centre and radius twice N0; produce line at right angles to NO till it meets the circle.)

PROBLEM (Euclid I. 12). Repeat.-The definition of a circle, and of a right angle: and the enunciation of Euc. I. 8 (page 46). General Enunciation. To draw a straight line perpendicular to a

given straight line from a given point

without it.
Particular Enunciation.
Given.—The straight line

A B and the point C.
Required.—To draw from Ca Ā

line perpendicular to A B. Construction. Take any point D on the side of A B remote

from C At the centre C and distance CD describe the

circle FDG, cutting A B in FG. (The whole circle is not shown in the figure, only the part

needed.) (a) Bisect FG (Euc. I. Io shows how), and call

the point of bisection H. Join CH The straight line CH drawn from the given point Cis perpendicular to the given straight

line A B. If CH is perpendicular to A B, the angle CHF

will be equal to the angle CHG.

Then we must prove that angle CHF is equal

to angle CHG. Proof (with syllogisms in full). Additional construction required for the purposes or

proof. Join CF, CG

2nd Syllogism.
If two triangles have two sides of the one

equal to two sides of the other, each to each,
and have also the base of one triangle, equal to
the base of the other, then those sides shall

contain equal angles. (Euc. I. 8.)
The two triangles CFH, CG H have the two

sides FH, HC equal to the two sides
GH, FC (c), and the base C F equal to the

base CHF (6).
(d) ... The angle CHF is equal to the

angle CHG
3rd Syllogism.
When a straight line standing on another straight

line makes the adjacent angles equal to
one another, the straight line which stands
on the other is called a perpendicular to it

(definition).
CH, standing on A B, makes the angle CHF

equal to the adjacent angle CHG (d). .. C H is perpendicular to A B.

Result.-Wherefore a perpendicular CH has been drawn to the given line A B from the given point C.

Q. E. F. EXERCISES.—I. Write out the proof of this proposition, omitting the major premiss of each syllogism, and giving the definition or proposition of Euclid referred to as the authority for the minor premiss. [Thus the first syllogism will read: Because C F and C G are drawn from the centre C to the circumference FDG... CF is equal to CG. (Euc. Def. of circle.)]