to the sum of the angles BAC, CAD, and DAG, together with the sum of the angles GAE, EAF, and FAB, and each of these sums is equal to two right angles by the last corollary.

DEFINITIONS.

21.-An angle less than a right angle is called an acute angle, and an angle greater than a right angle is called an obtuse angle.

22.-The angle by which any given angle falls short of two right angles is called the supplement of the given angle.

23. The angle by which any given angle falls short of one right angle is called the complement of the given angle.

24. If two straight lines, as AB and CD, cut each other in the point E, the angles AEC and BED are said to be vertically opposite angles, and so are the angles AED and B CEB (Fig. 33).

PROPOSITION 12.

If two straight lines cut one another, the vertically opposite angles shall be equal.

Let the straight lines AB and CD cut one another in the point E, then the vertically opposite angles AEC and BED shall be equal to each other, as also the angles AED and BEC.

Because AE stands upon CD,

therefore AEC+AED is equal to two right angles (Prop. 11). And because CE stands upon AB,

therefore AEC+ CEB is equal to two right angles (Prop. 11), therefore AEC+AED is equal to AEC+CEB.

Take away the common angle AEC,

therefore the angle AED is equal to the angle BEC. Similarly it may be proved that the angle AEC is equal to the angle BED.

PROPOSITION 13.

If at a point in a straight line two straight lines on opposite sides of it make the adjacent angles together equal to two right angles, or make the vertically opposite angles equal to one another, these two straight lines shall be in one and the same straight line.

to two right angles, then shall DC and CE be in the same straight line.

If CE be not in the same straight line with DC let, if possible, CF be in the same straight line with DC.

Because CA is drawn from the point C in the straight line DCF

therefore DCA+ACF is equal to two right angles (Prop. 11). But, by hypothesis,

DCA+ACE is equal to two right angles,

therefore DCA+ACF is equal to DCA+ACE. Take away the common angle DCA,

therefore the angle ACE is equal to the angle ACF
which is impossible.

Therefore CF is not in the same straight line with DC, and similarly it may be proved that no other line, except CE, is in the same straight line with DC,

therefore DCE is a straight line.

2nd. Let the two straight lines DC and CE on opposite sides of AB make the vertically opposite angles DCB and

ACE equal, then shall DC and CE be in the same straight line.

If not, let CF be in the same straight line with DC.

Because the straight lines ACB and DCF intersect in C, therefore the angle DCB is equal to the angle ACF (Prop. 12). But, by hypothesis,

the angle DCB is equal to the angle ACE.

Therefore the angle ACE is equal to the angle ACF, which is impossible.

Therefore CF is not in the same straight line with DC, and similarly it may be proved that no other line than CE can be in the same straight line with DC, therefore DC and CE are in the same straight line.

PROPOSITION 14.

The exterior angle, formed by producing any one side of a triangle, is greater than either of the interior and non-adjacent angles, and any two angles of a triangle are together less than two right angles.

E

B

C

D

Ist. The exterior angle ACD shall be greater than either of the interior non-adjacent angles BAC or ABC.

Let E be the middle point of AC, join BE and produce

it to F, making EF equal to BE, and join CF.

Because the two sides AE and EB, and the angle AEB, are respectively equal to the two sides CE and EF, and the angle CEF (Prop. 12 and construction),

therefore the triangles AEB and CEF are equal in all

their parts,

therefore the angle ACF is equal to the angle BAC,

therefore the angle ACD is greater than the angle BAC.

If AC be produced to G, it may be proved in a similar manner that the angle BCG is greater than the angle ABC. But the angle BCG is equal to the angle ACD (Prop. 12), therefore the angle ACD is greater than either of the

angles BAC or ABC.

2nd. Any two angles of the triangle, as ABC and ACB, shall be together less than two right angles.

Because the angle ACD is greater than the angle ABC, therefore ACD+ACB is greater than ABC+ACB. But ACD + ACB is equal to two right angles (Prop. 11), therefore ABC+ACB is less than two right angles.

PROPOSITION 15.

If a point be taken outside a given straight line, then— 1. It shall be always possible to draw one line, and only one line, from this point perpendicular to the given line, and lines equally inclined to this perpendicular shall be equal to one another.

2. The perpendicular shall be the shortest line that can be drawn from the given point to the given line, and of any pair of lines unequally inclined to the perpendicular the line nearer to it shall be less than that more remote.

Let AB be the given straight line, and C the point without it.

Fig. 37.

contains the point C and the line AB to be turned about the line AB until the portion of the plane above AB is brought into coincidence with the portion of the same plane below

AB, and let D be the point with which the point C coincides in this case.

Join CD, meeting AB in E, take F, any other point in AB, and join CF and DF.

Because the straight line EC is made to coincide with the straight line ED, while AB remains fixed,

therefore the angle AEC is equal to the angle AED, therefore each of the angles AEC and AED is a right angle.

But CEB is equal to AED (Prop. 12), therefore CEB and AEC are each right angles, and EC is perpendicular to AB.

Also no other line except CE can be drawn from Cat right angles to AB.

For, if it were so drawn, we should have two angles of a triangle together equal to two right angles, which is impossible, by Prop. 14.

Let CG be a line from C, such that the angle ECG is equal to the angle ECF.

Then in the triangles ECG and ECF the two angles FEC and FCE of the one are equal to the two angles GEC and GCE of the other, each to each, and the side EC adjacent to the equal angles is common;

therefore the triangles are equal in all their parts, and the side CG is equal to the side CF.

2nd. CE is the shortest line from C to AB, and CF which is nearer to CE, is less than CH, which is more remote. Because CEF is a right angle, therefore CFE is less than a right angle (Prop. 14),

therefore CFE is less than CEF,

therefore CE is less than CF (Prop. 8),

i.e. CE is less than any other straight line from C to AB. For the same reason as above the angle CHF is less than a right angle.

Also the angle CFH is greater than the interior angle CEF, and therefore greater than a right angle (Prop. 14), therefore CFH is greater than CHF,

therefore CF is less than CH (Prop. 8).