Equal triangles upon the same base, or upon equal bases in the same straight line and on the same sides of them, are between the same parallels.

B

Let the triangles ABC, BDC be upon the same base BC, and on the same side of it, they are between the same parallels. Join AD, if it is not parallel to BC, the parallel must either lie below or above AD, in the positions AE or AF; but the parallel cannot lie below AD, in the position AE, because then two straight lines which intersect would be both parallel to the same straight line, but this is (39, Cor.) impossible; therefore the parallel cannot lie below AD. For the same reason, it cannot lie above AD, in the position AF; therefore since B the parallel can neither lie below or above AD, the parallel must lie upon it; wherefore AD is parallel to BC.

A

C E

Next, let ABC, DEF be equal triangles upon equal bases BC, EF, in the same straight line, they are between the same parallels. For if AD be joined, and the lines AG and AH drawn, it may be demonstrated, as before, that AD is parallel to BF.

the

Cor. Equal triangles between the same parallels must stand on equal bases. For if the base BC were greater than EF, triangle ABC would (Ax. 8) be greater than DEF; and if the base BC were less than EF, the triangle ABC would be less than DEF; and therefore, in order that the triangles may be equal, their bases must be equal.

PROPOSITION XLVII. THEOR.

If a parallelogram and a triangle are upon the same base and between the same parallels, the parallelogram is double the triangle.

the

DE

Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE, parallelogram ABCD is double the triangle EBC. Join AC, then the triangle ABC is equal A (44, Cor.) to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double (40) the triangle ABC, because the diagonal AC divides it into two equal parts; wherefore the parallelogram ABCD is double the triangle EBC.

B

C

PROPOSITION XLVIII. THEOR.

The complements of the parallelograms which are about the diagonal of any parallelogram are equal.

Let ABCD be a parallelogram of which the diagonal is AC; let EH and FG be the parallelograms about AC, that is, through which AC passes, and let BK and KD be the other parallelograms which make up the whole figure ABCD, and are therefore called the complements, the complement BK is equal to the complement KD.

E

A H

D

K

F

Because ABCD is a parallelogram and AC its diagonal, the triangle ABC is equal (40) to the triangle ADC; and because EKHA is a parallelogram and AK its diagonal, the triangle AEK is equal to AHK; and, for the B Ꮐ same reason, the triangle KGC is equal to KFC; and by adding (Ax. 2), these equals, the triangles AEK, KGC are

equal to the triangles AHK, KFC. But the whole triangle ABC is equal to the whole ADC; therefore (Ax. 3) the remaining complement BK is equal to the remaining complement KD.

PROPOSITION XLIX. THEOR.

All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides wanting four right angles.

Let ABCDE be any given rectilineal figure, all its interior angles are equal to twice as many right angles as the figure has sides wanting four right angles.

E

A

D

C

B

From F, a point within the figure, draw the straight lines AF, BF, CF, DF, EF, to its angular points, the figure ABCDE is divided into as many triangles as the figure has sides. Now all the angles of these triangles are equal to twice as many right angles (30) as there are sides of the figure, and the same angles are equal to the angles of the figure together, with the angles at the point F, which is the common vertex of the triangles, that is (7, Cor. 2), together with four right angles; therefore twice as many right angles as the figure has sides are equal to all the angles of the figure together with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides wanting four right angles.

Cor. 1. All the interior angles of a quadrilateral are therefore equal to four right angles, those of a pentagon to six right angles, etc., the sum increasing by two right angles for every additional side.

Cor. 2. The quadrilateral is the only figure which has the sum of its interior and exterior angles equal.

PROPOSITION L. THEOR.

All the exterior angles of any rectilineal figure are together equal to four right angles.

Let ABCDE be a rectilineal figure, and let its sides be produced, all the exterior angles are together equal to four right angles.

Because every interior angle as ABC, with its adjacent exterior angle CBF, is equal (5) to two right angles; therefore the interior, E together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides; that is, by the last proposition, equal to all the interior angles of the figure together with four right angles; therefore all the exterior angles are together equal to four right angles.

A

B F

Scholium. If the figure has what is called a re-entrant angle, its exterior angles shall be as much greater than four right angles as the re-entrant angle is less than two right angles.

PROPOSITION LI. THEOR.

If two sides of a triangle be bisected, the straight line joining the points of section is parallel to the remaining side of the triangle, and equal to the half of it.

Let ABC be a triangle, and let its two sides AB, AC be bisected in D and E, the straight line joining

D and E is parallel to the side BC, and equal to the half of it.

Join DE, and produce it to F, and make EF D equal to DE, and join FC. In the triangles ADE, CEF the side AE is equal to CE, and the side DE to EF, and the included angles at B

A

E

E are (7) equal; the triangle ADE is therefore equal (12) to

E

CEF, and the side AD is equal to CF, but AD is equal to BD; therefore CF is equal to BD, and the angle ADE is equal to EFC, and these are alternate angles; therefore AB or BD is parallel to FC, and have their adjacent extremities joined by BC and DF, and therefore BC, DF are equal and parallel, and the figure BDFC is a parallelogram, and therefore its opposite sides DF and BC are equal, but DE is half of DF; wherefore DE is also equal to half of BC.

Cor. 1. If DE bisects AB, it will also bisect AC. For in the triangles ADE, CEF the sides AD, DE are equal to CF, FE, and the angles ADE and F are equal; therefore the triangles are equal, and AE is equal to EC; wherefore if a straight line, parallel to the base of a triangle, bisects one of the sides, it shall also bisect the other.

Cor. 2. By means of this proposition a triangle may be transformed into a parallelogram, and a parallelogram into a triangle having the same base and equal areas. The altitude of the parallelogram shall be half that of the triangle.

Cor. 3. If the triangle be equilateral, the parallelogram and the triangle have both equal areas and perimeters.

PROPOSITION LII. THEOR.

The straight line drawn from the middle of the base to the vertex of a triangle is greater, equal to, or less than half the base, according as the vertical angle is acute, right, or obtuse.

Let ABC be a triangle, and let the straight line AD be drawn from D, the middle point of the base to A, the vertex AD is greater, equal to, or less than the half of BC, according as the vertical angle A is acute, right, or obtuse.

CASE 1. Let the angle A be acute. Bisect AC in E, and join DE, the line DE is parallel (51) to AB, and the exterior angle DEC is equal (35) to the interior opposite angle BAC,