Page images
PDF
EPUB

PROPOSITION XXIV. THEOREM.

If two angles have their sides parallel and lying in the same direction, the two angles will be equal.

Let BAC and DEF be the two angles, having AB parallel to ED, and AC to EF; then will the angles be equal.

A

B

[ocr errors]

E

G

F

For, produce DE, if necessary, till it meets AC in G. Then, since EF is parallel to GC, the angle DEF is equal to DGC (Prop. XX. Cor. 3.); and since DG is parallel to AB, the angle DGC is equal to BAC; hence, the angle DEF is equal to BAC (Ax. 1.).

Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, and ED in the same direction with AB, is necessary, because if FE were produced towards H, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC would be together equal to two right angles. For, DEH+DEF is equal to two right angles (Prop. I.); but DEF is equal to BAC: hence, DEH + BAC is equal to two right angles.

PROPOSITION XXV. THEOREM.

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

B

E

4

A D

Let ABC be any triangle: then will the angle C+A+B be equal to two right angles. For, produce the side CA towards D, and at the point A, draw AE parallel to BC. Then, since AE, CB, are parallel, and CAD cuts them, the exterior angle DAE will be equal to its inte-C rior opposite one ACB (Prop. XX. Cor. 3.); in like manner, since AE, CB, are parallel, and AB cuts them, the alternate angles ABC, BAE, will be equal: hence the three angles of the triangle ABC make up the same sum as the three angles CAB, BAE, EAD; hence, the sum of the three angles is equal to two right angles (Prop. I.).

Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

Cor. 2. If two angles of one triangle a re respectively equal to two angles of another, the third angles will also be equal. and the two triangles will be mutually equiangular.

Cor. 3. In any triangle there can be but one right angle: for if there were two, the third angle must be nothing. Still less, can a triangle have more than one obtuse angle.

Cor. 4. In every right angled triangle, the sum of the two acute angles is equal to one right angle.

Cor. 5. Since every equilateral triangle is also equiangular (Prop. XI. Cor.), each of its angles will be equal to the third part of two right angles; so that, if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by 3.

Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE is equal to the angle C.

PROPOSITION XXVI. THEOREM.

The sum of all the interior angles of a polygon, is equal to two right angles, taken as many times less two, as the figure has sides.

G

Let ABCDEFG be the proposed polygon. If from the vertex of any one angle A, diagonals B AC, AD, AE, AF, be drawn to the vertices of all the opposite angles, it is plain that the polygon will be divided into five triangles, if it has seven sides; into six triangles, if it has eight; and, in general, into as many triangles, less two, as the polygon has sides; for, these triangles may be considered as having the point A for a common vertex, and for bases, the several sides of the polygon, excepting the two sides which form the angle A. It is evident, also, that the sum of all the angles in these triangles does not differ from the sum of all the angles in the polygon: hence the sum of all the angles of the polygon is equal to two right angles, taken as many times as there are triangles in the figure; in other words, as there are units in the number of sides diminished by two.

Cor. 1. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four

right angles: hence, if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanctions the seventeenth Definition, where the four angles of a quadrilateral are asserted to be right angles, in the case of the rectangle and the square.

Cor. 2. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles: hence, when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or to § of one right angle.

Cor. 3. The sum of the angles of a hexagon is equal to 2× (6—2,) or eight right angles; hence in the equiangular hexagon, each angle is the sixth part of eight right angles, or of one.

Scholium. When this proposition is applied to polygons which have re-entrant angles, each reentrant angle must be regarded as greater than two right angles. But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons

with salient angles, which might otherwise be named convex polygons. Every convex polygon is such that a straight line, drawn at pleasure, cannot meet the contour of the polygon in more than two points.

PROPOSITION XXVII. THEOREM.

If the sides of any polygon be produced out, in the same direction, the sum of the exterior angles will be equal to four right angles.

Let the sides of the polygon ABCDFG, be produced, in the same direction; then will the sum of the exterior angles a+b+c+d+f+g, be equal to four right angles.

A

a

[ocr errors]

D

B/b

For, each interior angle, plus its exterior angle, as A+a, is equal to two right angles (Prop. I.). But there are as many exterior as interior angles, and as many of each as there are sides of the polygon : hence, the sum of all the interior and exterior angles is equal to twice as many right angles as the polygon has sides. Again, the sum of all the interior angles is equal to two right angles, taken as many times, less two, as the polygon has sides (Prop. XXVI.); that is, equal to twice as many right angles as the figure has sides, wanting four right angles. Hence, the interior angles plus four right

angles, is equal to twice as many right angles as the polygon has sides, and consequently, equal to the sum of the interior angles plus the exterior angles. Taking from each the sum of the interior angles, and there remains the exterior angles, equal to four right angles.

PROPOSITION XXVIII. THEOREM.

In every parallelogram, the opposite sides and angles are equal.

D

B

Let ABCD be a parallelogram: then will AB=DC, AD=BC, A=C, and ADC=ABC. For, draw the diagonal BD. The triangles ABD, DBC, have a common side BD; and A since AD, BC, are parallel, they have also the angle ADB DBC, (Prop. XX. Cor. 2.); and since AB, CD, are parallel, the angle ABD BDC: hence the two triangles are equal (Prop. VI.); therefore the side AB, opposite the angle ADB, is equal to the side DC, opposite the equal angle DBC; and the third sides AD, BC, are equal: hence the opposite sides of a parallelogram are equal.

Again, since the triangles are equal, it follows that the angle A is equal to the angle C; and also that the angle ADC composed of the two ADB, BDC, is equal to ABC, composed of the two equal angles DBC, ABD: hence the opposite angles of a parallelogram are also equal.

Cor. Two parallels AB, CD, included between two other parallels AD, BC, are equal; and the diagonal DB divides the parallelogram into two equal triangles.

PROPOSITION XXIX. THEOREM.

If the opposite sides of a quadrilateral are equal, each to each, the equal sides will be parallel, and the figure will be a parallelogram.

Let ABCD be a quadrilateral, having its opposite sides respectively equal, viz. AB=DC, and AD=BC; then will these sides be parallel, and the figure be a parallelogram.

A

[blocks in formation]

B

For, having drawn the diagonal BD, the triangles ABD, BDC, have all the sides of the one equal to

the corresponding sides of the other; therefore they are equal, and the angle ADB, opposite the side AB, is equal to DBC, opposite CD (Prop. X.); therefore, the side AD is parallel to BC (Prop. XIX. Cor. 1.). For a like reason AB is parallel to CD therefore the quadrilateral ABCD is a parallelogram.

PROPOSITION XXX. THEOREM.

If two opposite sides of a quadrilateral are equal and parallel, the remaining sides will also be equal and parallel, and the figure will be a parallelogram.

Let ABCD be a quadrilateral, having D the sides AB, CD, equal and parallel; then will the figure be a parallelogram.

A

B

C

For, draw the diagonal DB, dividing the quadrilateral into two triangles. Then, since AB is parallel to DC, the alternate angles ABD, BDC, are equal (Prop. XX. Cor. 2.); moreover, the side DB is common, and the side AB=DC; hence the triangle ABD is equal to the triangle DBC (Prop. V.); therefore, the side AD is equal to BC, the angle ADB DBC, and consequently AD is parallel to BC; hence the figure ABCD is a parallelogram.

PROPOSITION XXXI. THEOREM.

The two diagonals of a parallelogram divide each other into equal parts, or mutually bisect each other.

Let ABCD be a parallelogram, AC and B DB its diagonals, intersecting at E, then will AE EC, and DE EB.

Comparing the triangles ADE, CEB, we find the side AD=CB (Prop. XXVIII.), the angle ADE=CBE, and the angle

E

DAE ECB (Prop. XX. Cor. 2.); hence those triangles are equal (Prop. VI.); hence, AE, the side opposite the angle ADE, is equal to EC, opposite EBC; hence also DE is equal to EB.

Scholium. In the case of the rhombus, the sides AB, BC, being equal, the triangles AEB, EBC, have all the sides of the one equal to the corresponding sides of the other, and are therefore equal: whence it follows that the angles AEB, BEC, are equal, and therefore, that the two diagonals of a rhombus cut each other at right angles.

с

« PreviousContinue »