PRACTICAL GEOMETRY.

PROBLEMS.

The following problems are to be solved with ruler and compasses only. No step requires the actual measurement of any line or angle; that is to say, the constructions are to be made without using either a graduated scale of length, or a protractor.

The problems are not merely to be studied as propositions ; but the construction in every case is to be actually performed by the learner, great care being given to accuracy of drawing.

Each problem is followed by a theoretical proof; but the results of the work should always be verified by measurement, as a test of correct drawing. Accurate measurement is also required in applications of the problems.

In the diagrams of the problems lines which are inserted only for purposes of proof are dotted, to distinguish them from lines necessary to the construction.

For practical applications of the problems the student should be provided with the following instruments:

1. A flat ruler, one edge being graduated in centimetres and millimetres, and the other in inches and tenths.

2. Two set squares; one with angles of 45°, and the other with angles of 60° and 30°.

3. A pair of pencil compasses.

4.

5.

A pair of dividers, preferably with screw adjustment.
A semi-circular protractor.

PROBLEM 1.

To bisect a given angle.

Let BAC be the given angle to be bisected. Construction. With centre A, and any radius, draw an arc of a circle cutting AB, AC at P and Q.

With centres P and Q, and radius PQ, draw two arcs cutting
Join AO.
BAC is bisected by AO.

at O.

Then the

.. the triangles are equal in all respects; Theor. 7.

so that the PAO the QAO;

=

that is, the BAC is bisected by AO.

NOTE. PQ has been taken as the radius of the arcs drawn from the centres P and Q, and the intersection of these arcs determines the point O. Any radius, however, may be used instead of PQ, provided that it is great enough to secure the intersection of the arcs.

PROBLEM 2.

To bisect a given straight line.

B

Let AB be the line to be bisected.

Construction. With centre A, and radius AB, draw two arcs, one on each side of AB.

With centre B, and radius BA, draw two arcs, one on each side of AB, cutting the first arcs at P and Q.

Proof.

because

Join PQ, cutting AB at O.

Then AB is bisected at O

Join AP, AQ, BP, BQ.

In the ▲ APQ, BPQ,

AP BP, being radii of equal circles,
AQ=BQ, for the same reason,

and PQ is common;

NOTES. (i) AB was taken as the radius of the arcs drawn from the centres A and B, but any radius may be used provided that it is great enough to secure the intersection of the arcs which determine the points P and Q.

(ii) From the congruence of the AAPO, BPO it follows that the LAOP the BOP. As these are adjacent angles, it follows that PQ

bisects AB at right angles.

PROBLEM 3.

To draw a straight line perpendicular to a given straight line at a given point in it.

Let AB be the straight line, and X the point in it at which a perpendicular is to be drawn.

Construction. With centre X cut off from AB any two equal parts XP, XQ.

With centres P and Q, and radius PQ, draw two arcs cutting

And these being adjacent angles, each is a right angle ;

that is, XO is perp. to AB.

Obs. If the point X is near one end of AB, one or other of the alternative constructions on the next page should be used