Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/310

286 be that of AC to CE. Join EB; through C draw a straight line parallel to EB; then this straight line will meet AB, produced through B, at the required point.

VI. II. This is a particular case of VI. 12.

VI. 14. The following is a full exhibition of the steps which lead to the result that FB and BG are in one straight line.

The angle DBF is equal to the angle GBE; [Hypothesis.

add to each the angle FBE;

therefore the angles DBF, FBE are together equal to the angles GBE, FBE. [Axiom 2.

But the angles DBF, FBE are together equal to two right angles; [I. 13.

therefore the angles GBE, FBE are together equal to two right angles; [Axiom 1.

therefore FB and BG are in one straight line. [I. 14.

VI. 15. This may be inferred from VI. 14, since a triangle is half of a parallelogram, with the same base and altitude.

It is not difficult to establish a third proposition conversely connected with the two involved in VI. 14, and a third proposition similarly conversely connected, with the two involved in VI. 15. These propositions are the following.

Equal parallelograms wJiich have their sides reciprocally proportional, have their angles equal, each to each. Equal triangles which have the sides about a pair of angles reciprocally proportional, have those angles equal or together equal to two right angles.

We will take the latter proposition.

Let ABC, ADE be equal triangles; and let CA be to AD as AE is to AB: either the angle BAC shall be equal to the angle DAE, or the angles BAC and DAE shall be together equal to two right angles.

[The student can construct the figure for himself. ]

Place the triangles so that CA and AD may be in one straight line; then if EA and AB are in one straight line the angle BAC is equal to the angle DAE. [I. 15.

If EA and AB are not in one straight line, produce BA through A to F, so that AF may be equal to AE; join DF and EF.

Then because CA is to AD as AE is to AB, [Hypothesis.

and AF is equal to AE, [Construction.

therefore CA is to AD as AF is to AB. [V. 9, V.11

Therefore the triangle DAF is equal to the triangle BAC. [VI. 15.