THEOREM XV.

In any triangle the sum of all the three angles is equal to two right angles.

Let ABC be any plane triangle ; then the sum of the three angles A+ B+C is equal to two right angles.

For, let the side AB be produced to D. Then the outward angle CBD A

C

B D

is equal to the sum of the two inward opposite angles AC (th. 13). To each of these equals add the inward angle B; then will the sum of the three inward angles A+B+C be equal to the sum of the two adjacent angles ABC + CBD (ax. 2). But the sum of these two last adjacent angles is equal to two right angles (th. 6). Therefore, also, the sum of the three angles of the triangle A+B+C is equal to two right angles (ax. 1). Q. E. D.

Corol. 1. If two angles in one triangle be equal to two angles in another triangle, the third angles will also be equal; for they make up two right angles in both.

Corol. 2. Two right-angled triangles will be equiangular when they have an acute angle in each equal. Corol. 3. The sum of two angles of any triangle and the third angle are supplements of each other.

Corol. 4. If one angle in one triangle be equal to one angle in another, the sums of the remaining angles will also be equal (ax. 3).

Corol. 5. If one angle of a triangle be right, the sum of the other two will also be equal to a right angle, and each of them singly will be acute, or less than a right angle, and will be the complement of the other.

Corol. 6. The two least angles of every triangle are acute, or each less than a right angle. In other words, there can be but one right angle, or one obtuse angle in a triangle.

Corol. 7. Any two angles and a side of one trian

gle being equal to the same in another, the triangles will be equal (th. 2).

Corol. 8. One side and an acute angle of a rightangled triangle being equal to the same in another, the triangles are equal.

THEOREM XVI.

The sum of all the inward angles of a polygon is equal to twice as many right angles, wanting four, as the figure has sides.

Let ABCDE be any figure; then the sum of all its inward angles, A + B + C + D + E, is equal to twice as many right angles, wanting four, as the figure has sides.

E

A

D

B

For, from any point F, within it, draw lines, FA, FB, FC, &c., to all the angles, dividing the polygon into as many triangles as it has sides. Now the sum of the three angles of each of these triangles is equal to two right angles (th. 15); therefore, the sum of the angles of all the triangles is equal to twice as many right angles as the figure has sides. But the sum of all the angles about the point F, which are so many of the angles of the triangles, but no part of the inward angles of the polygon, is equal to four right angles (corol. 3, th. 6), and must be deducted out of the former sum. Hence it follows that the sum of all the inward angles of the polygon alone, A+B+C+D + E, is equal to twice as many right angles as the figure has sides, wanting the said four right angles. Q. E. D.

Corol. 1. In any quadrangle, the sum of all the four inward angles is equal to four right angles.

Corol. 2. Hence, if three of the angles be right ones, the fourth will also be a right angle.

Corol. 3. And if the sum of two of the four angles

be equal to two right angles, the sum of the remaining two will also be equal to two right angles.

Corol. 4. The sum of the angles of a pentagon is 5 × 2—4=6 right angles. Of a hexagon, 6 x 2× 4-8 right angles.* Of a polygon of n sides (n × 2— 4) right angles.

The rule may be thus expressed to obtain the sum of the angles of a polygon, double the number of sides and subtract 4, the right angle being the unit of measure.†

THEOREM XVII.

A perpendicular is the shortest line that can be drawn from a given point to an indefinite line. And, of any other lines drawn from the same point, those that are equally distant from the perpendicular are equal, and those nearest the perpendicular are less than

those more remote.

If AB, AC, AD, &c., be lines drawn from the given point A, to the indefinite line DE, of which AB is perpendicular; then shall the perpendicular AB be less than AC, AE AC if BE BC, and AC<AD if BC<BD. For, the angle B being a right one, the angle C of

==

D

E

B

* Each angle of a regular hexagon would be the sixth part of 8 right angles, or or or 1 right angles, i. e., 120°, 90° being a right angle, this is of 360°, or 4 right angles; so that if three regular hexagons were placed together they would fill up the whole angular space about a point.

This would not be the case with pentagons, for 5 × 2-4

5

=1, which, multiplied by 3, gives 33 < 4 right angles, and, multiplied by 4, gives 44 > 4 right angles, so that they would fit together neither by threes nor fours.

It will be found, on examination, that no other figures except squares and triangles have this same property with the hexagon. Hence these three kinds of figures alone are employed for paving blocks. + The above applies only to convex polygons, that is, those in which the angles point outward. No convex polygon can have more than three acute angles.

the triangle ABC is acute (by cor. 6, th. 15), and therefore less than the angle B. But the less angle of a triangle is subtended by the less side (th. 9). Therefore the side AB is less than the side AC.

Again, in the right-angled triangles ABC, ABE the two sides AB, BC being respectively equal to the two AB, BE, the third sides are equal (th. 1).

Corol. Every point, as A, of a perpendicular at the middle of a given line CE is equally distant from its extremities C and E.

Finally, the angle ACB being acute, or less than a right angle, as before, the adjacent angle ACD will be greater than a right angle, or obtuse (by th. 6); consequently, the angle D is acute (corol. 6, th. 15), and therefore is less than the angle C. And since the less side is opposite to the less angle, therefore the side AC is less than the side AD. Q. E. D.

Corol. The least distance of a given point from a line is the perpendicular. For if it were an oblique line the perpendicular would be shorter, and thus less than the least distance, which is impossible.

THEOREM XVIII.

Every point out of a perpendicular at the middle of a given line is at unequal distances from the extremities of the line.

Let DC be a perpendicular at the middle of AB, and I a point out of the perpendicular, then shall IB < IA. For join BD; then, BI < BD + DI ; or since BD = AD (th. 17), BI < AD +DI, or BI < AI. Q. E. D.

Scholium. There can be but one perpendicular through a given point to a given line. For there can be

D

B

E

but one line through the same point in the same direction, or having the same difference of direction from a given line.

B

THEOREM XIX.

The opposite sides and angles of a parallelogram are equal to each other, and the diagonal divides it into two equal triangles.

Let ABDC be a parallelogram, of which the diagonal is BC; then will its opposite sides and angles be equal to each other, and the diagonal BC will divide it into two equal parts, or triangles.

A

B

For, since the sides AB and DC are parallel, as also the sides AC and BD (def. 30), and the line BC meets them; therefore the alternate angles are equal (th. 10), namely, the angle ABC to the angle BCD, and the angle ACB to the angle CBD. Hence the two triangles, having two angles in the one equal to two angles in the other, have also their third angles equal (cor. 1, th. 15), namely, the angle A equal to the angle D, which are two of the opposite angles of the parallelogram.

Also, if to the equal angles ABC, BCD be added the equal angles CBD, ACB, the wholes will be equal (ax. 2), namely, the whole angle ABD to the whole ACD, which are the other two opposite angles of the parallelogram. QE.D.

Again, since the two triangles are mutually equiangular, and have a side in each equal, viz., the common side BC; therefore the two triangles are identical (th. 2), or equal in all respects, namely, the side AB equal to the opposite side DC, and AC equal to the opposite side BD,* and the whole triangle ABC equal to the whole triangle BCD. Q. E. D.

Corol. 1. Hence, if one angle of a parallelogram be a right angle, all the other three will also be right

*The student will observe that in identical triangles the equal sides are opposite equal angles. Thus, in the diagram, the side BD opposite the angle C, in the lower triangle, is equal to the side AC, opposite the equal angle B in the upper.