PREFACE.

THE following pages contain solutions of the Exercises given in the Explicit Euclid in CHAMBERS'S EDUCATIONAL Course, forming a KEY to that volume. Such a Key will be useful to those privately prosecuting the study of Geometry; and when consulted only after the Student has made every effort from his own knowledge to solve an Exercise, will be found to supply, in some measure, the office of an Instructor. It will likewise be of advantage to those Teachers whose school-room duties are so varied and miscellaneous as to prevent them from bestowing that time and study which Mathematical Solutions in general require. Nor will the volume be without its use to every class of Geometrical Scholars, for it may be read as a sequel to the ordinary course of study contained in the Elements of Euclid; and in its pages will be found the solution of many Geometrical Propositions not contained in Euclid's Elements, the knowledge of which will facilitate the student's progress in the study of Natural Philosophy, and other branches of Applied Mathematics.

Several of the Exercises given are original; and in the Solutions of all, the Author has endeavoured to be as plain and explicit as possible-adopting methods more simple and concise than are to be found in most collections of a similar nature, but, at the same time, giving all the demonstrations entire.

EDINBURGH, October 1865.

KEY TO PLANE GEOMETRY

EXPLICITLY ENUNCIATED.

FIRST BOOK.

EXERCISE I.-THEOREM.

A LINE that bisects the vertical angle of an isosceles triangle, also bisects the base perpendicularly.

Given ABC an isosceles triangle, of which the vertical angle ACB is bisected by the line CD, to prove that

AB is bisected in D, and that CD is perpendicular to AB.

=

C

For AC CB (I. Def. 27), and CD is common to the two triangles ACD, BCD; therefore the two sides, AC, CD, are equal to BC, CD, each to each, and the contained angle ACD is equal to the contained angle BCD (by hypothesis); hence the triangles are equal in all respects (I. 4), therefore AD = BD, and the angle ADC = BDC; but they are adjacent angles; therefore each of them is a right angle, hence CD is perpendicular to AB (I. Def. 11).

A D

B

EXERCISE II.-THEOREM.

If a line be bisected perpendicularly by another line, every point in the latter is equally distant from the extremities of the former; and any point not situated in the latter is at unequal distances from the extremities of the former.

Let EF be bisected perpendicularly by PL, then any point G in PL is equidistant from E and F; and a point K, not situated in PL, is at unequal distances from E and F.

For join GE, and GF; then, because EI = IF (by hypothesis),

A