Cor. 4. If one angle in any triangle be a right angle, then the other two angles together are equal to a right angle. ED.

Note. Two such angles are called complements of each other to a right angle.

Cor. 5. Each angle of an equilateral triangle is one third of two right angles, or two thirds of one right angle.

Let A denote each angle, then 3 A-2 right angles, therefore A = of 2 right angles, or of one right angle.

ED.

Cor.. 6. In every quadrilateral figure

B

the sum of the four angles is equal to A four right angles.

For if a line BC be drawn between two opposite angles of any quadrilateral figure ACDB, it will divide the figure into two triangles ABC, BCD. Now the sum of the three angles of each triangle is equal to two right angles; therefore all the angles of both triangles, which make the four angles of the quadrilateral figure, are together equal to four right angles.

C

PROPOSITION XXXIII. THEOREM.

D

ED.

The two straight lines which join the corresponding extremities of two equal and parallel straight lines are also equal and parallel.

Let AB, CD be two equal

and parallel lines, and let the A lines AC, BD join their corresponding extremities; then AC, BD are also equal and parallel.

Draw the diagonal BC. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are

C

B

equal (Prop. 29). Because the side AB is equal to CD, and BC is common to the two triangles ABC, DCB, and the angle ABC is equal to BCD, the base AC is equal to BD (Prop. 4.), and the angle ACB to CBD. Because the line BC meets the two lines AC, BD, and makes the alternate angles ACB, CBD equal to each other, AC is parallel to BD (Prop. 27.); and AC was shown to be equal to BD. Therefore, the two straight lines &c. Q. E. D.

PROPOSITION XXXIV. THEOREM.

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

Let ACDB be a parallelogram, of which BC is a diagonal; the opposite sides of the figure are equal to each other, and also the opposite angles; and the diagonal BC bisects it.

Because AB is parallel to CD, A

B

and BC meets them, the alternate angles ABC, BCD are equal to each other (Prop. 29); and beeause AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to each other; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one equal to two agles BCD, CBD in the other, each to each, therefore the remaining angles BAC, BDC are equal (2 Cor. 32.), and the whole angles ABD, ACD are equal. Hence the opposite angles of a paral. are equal.

C

D

Again, the triangles ABC, BCD are mutually equiangular, and the side BC is common to both, therefore they are equal in all respects (A), and the side AB is equal to CD, and the side AC to BD, and the triangle ABC to BCD. Therefore the opposite sides of a paral. are equal, and the diagonal bisects it. Wherefore the opposite sides &c. Q. E. D.

PROPOSITION B. THEOREM.

ED.

If the opposite sides of a quadrilateral figure be equal, they are also parallel, that is, the figure is a parallelogram.

Let ABDC be a quadrilateral figure, having its opposite sides equal, namely, AB equal to CD,

and AC equal to BD; the opposite

sides are parallel.

A

B

Draw the diagonal BC; then, because the side AB is equal to CD), and the side AC to BD, and the side BC is common to the triangles ACB, BCD, the angles ABC, BCD are equal (Prop. 8.) and the angles ACB, CBD are equal. But ABC, BCD are alternate angles; there

C

D

fore AB, CD are parallel (Prop. 27.). Also ACB, CBD are alternate angles; therefore AC, BD are parallel. Hence ABDC is a parallelogram. Therefore, if the opposite sides &c. Q. E. D.

COR. If the opposite sides of a quadrilateral figure be equal, the opposite angles are also equal. For the figure is a parallelogram, therefore the opposite angles are equal.

PROPOSITION XXXV. THEOREM.

Parallelograms on the same base, and between the same parallels, are equal to one another; that is, their surfaces are equal.

Let the parallelograms ABCD, DBCF be on the same base BC, and between the same parallels AF, BC; they are equal in surface.

If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, it is plain that each of the parallelograms is double of the triangle BDC (Prop. 34); therefore they are equal to each other.

A

B

D

F

C

But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same. point; then, because ABCD is a paral. AD is equal to BC (Prop. 34). For the same reason EF is equal to BC. Wherefore AD is equal to EF. From the whole line AF take the two equal parts AD, EF, and the remainders DF, AE will be equal.

Now AB is equal to DC; and the exterior angle FDC is equal to the interior angle EAB (Prop. 29); wherefore the tri

angle EAB is equal to the triangle FDC (Prop. 4). Take the triangle FDC from the trapezium ABCF, and the paral. ABCD

will remain; and take the triangle AEB from the trap. ABCF, and the paral. EBCF will remain. Therefore the paral. ABCD is equal to EBCF. Therefore, parallelograms &c. Q. E. D.

PROPOSITION XXXVI. THEOREM.

Parallelograms on equal bases, and between the same parallels, are equal to one another.

cause BC is equal to B FG, and FG to EH

(Prop. 34), BC is equal

A

to EH. But BC, EH are parallels, and joined at their corresponding extremities by the lines BF, CH; therefore BE, CH are equal and parallel (Prop. 33); therefore EBCH is a paral. and it is equal to ABCD (Prop. 35). But the paral. EFGH is equal to EBCH; therefore also the paral. ABCD is equal to EFGH. Wherefore, parallelograms &c. Q. E. D.

Otherwise. This is an obvious deduction from the last proposition, and needs no formal proof by a figure.

Suppose the base of one paral. to be applied to the base of the other; then the bases will coincide, because they are equal, and the two parals. will stand on the same base and between the same parallels. Therefore they are equal to each other. Therefore, parallelograms &c. Q. E. D.

PROPOSITION XXXVII. THEOREM.

ED...

Triangles on the same base, and between the same parallels, are equal to one another; that is, their surfaces are equal.

Let the triangles ABC, DBC be on the same base BC, and between the same parallels AD,BC; the triangle ABC is equal to DBC.

Produce AD both ways to the points E, F; through B draw BE parallel to CA (Prop. 31), and through C draw CF parallel to BD; then each of the figures EBCA, DBCF is a paral.; and EBCA is equal to DBCF (Prop. 35).

E

A D

F

C

Now the triangle ABC is half of the paral. EBCA (Prop. 34); and the triangle DBC is half of the paral. DBCF; therefore the triangle ABC is equal to DBC (7 ̊Ax.). Wherefore, triangles &c. Q. E. D.

Otherwise. This is a cor. to prop. 35. For triangles are the halves of parallelograms on the same base and between the same parallels (Prop. 34.), and therefore are equal to one another.

PROPOSITION XXXVIII. THEOREM.

ED.

Triangles on equal bases, and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD; the triangle ABC is equal to DEF.

Produce AD both ways to the points G, H; through B draw BG parallel to CA (Prop. 31), and through F draw FH parallel

to ED; then the figures GBCA, DEFH are parallelograms; and they are equal (Prop. 36).

Now the triangle ABC is half of the paral. GBCA (Prop. 34), and the triangle DEF is half of the paral. BEFH; therefore the triangle ABC is equal to DEF (7 Ax.). Wherefore, triangles &e. Q. E. D.

Otherwise. This is a cor. to prop. 36. For they are the halves of parals. on equal bases and between the same parallels, and therefore are equal.

ED.