A B angles BCD, CBD in the other, each to each, and one side BC, common to the two triangles; therefore the two triangles are equivalent (6. 2.), and their other sides are equal, each to each; and the third angle of the one is equal to the third angle of the other, viz. the side AB to the side CD, and AC to BD, and the angle BAC is equal to the angle BDC; and because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; therefore the whole angle ABD is equal to the whole angle ACD; and the angle BAC has been shown to be equal to the angle BDC. Therefore, &c. D COR. Because the triangles ABC, BCD are equivalent, as proved above; therefore the diagonal bisects the parallelogram. PROP. X. THEOR. Two straight lines which are at the same distance at each extremity, are parallel. Let the straight lines AB and CD be at the same distance at A as at B; they shall be parallel. From A draw AC perpendicular to CD, and from B, BD also per pendicular to CD. Then in the quadrangle ABDC, the sides AC and BD are equal by the hypothesis, and parallel, because they are both perpendicular (2 Cor. 2. 2.) to CD; therefore (7. 2.) the remaining sides AB and CD, are also equal and parallel. Q. E. D. PROP. XI. THEOR. The perpendiculars between parallel lines are equal. Let AB and FD be parallel lines, and AC and BD perpendiculars between them; AC is equal to BD. Because the straight line FD falls on the straight lines AC, BD, making the corresponding angles ACF, BDC, equal (1 Cor. 2. 2.), being both right angles (2 Cor. 1. 1.), AC is parallel to BD, and ABDC is a parallelogram; and therefore AC is equal (9. 2.) to BD. Therefore the perpendiculars between parallel lines are equal. Which was to be proved. COR. If parallel lines are produced indefinitely from either extremity, they remain always at the same distance from each other. If two lines make an angle, and two points be taken in one of the lines, so that one of the points is twice the distance of the other from the angu lar point, it shall also be twice the distance from the other line. Let the straight lines AB and AD make an angle, and let the points C and D be taken in AD, so that AD shall be double of AC, and CE and BD be drawn perpendicular to EB; BD shall be double of CE. Produce CE to F, make CF equal to CE, and join FD. Because the two triangles ACE, FDC have the two sides AC, CE equal to the two sides CD, CF, each to each; and likewise have the contained angle ACE equal (3. 1.) to the contained angle FCD, being vertical angles, therefore the triangles are equivalent (4. 1.), and B A E G F D H the angle CDF is equal to the angle CAE; and because the straight line AD falls on the straight lines AB, DF, making the angle DAE or CAE, equal to the corresponding angle ADF or CDF; therefore, FD is parallel (1 Cor. 2. 2.) to AB or BE; and because EF and BD are both perpendicular to BE, the angles FEB, DBE are each right angles, and are together equal to two right angles; therefore EF, BD are parallel (2 Cor. 2. 2.); and therefore BEFD is a parallelogram, and BD is equal (9. 2.) to EF; and because CF was made equal to CE, EF is double of CE, and therefore BD is also double of CE. Therefore, if two lines make an angle, and two points be taken in one of the lines, so that one of the points is twice the distance of the other from the angular point, it shall also be twice the distance from the other line. Which was to be proved. COR. Produce AB and AD to G and H, so that AH shall be double of AD, and draw GH perpendicular to AG: it may be proved in the same manner, that GH is double of BD, and if they are continually produced, the perpendiculars will be continually increased; and therefore they may be produced until the perpendicular is greater than any given straight line. PROP. XIII. THEOR. If a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet upon that side on which are the angles, which are less than two right angles. Let the straight line AC fall on the straight lines AB, BC, so as to make the angles BAC, ACB together less than two right angles; AB and CB, if continually produced, shall at length meet. Through C draw CD, parallel to AB; then because AB is parallel to CD, the angles BAC, ACD are together equal (1 Cor. 3. 2.) to two right angles; but the angles BAC, ACB are, by the hypothesis, together less than two right angles; and are therefore less than the angles BAC, ACD; take away the common angle BAC, and the remain A B C D ing angle ACB is less than the remaining angle ACD; and the angle ACD is composed of the angles ACB, BCD; and because the straight lines BC, CD make an angle, if they are produced, the distance between their extremities will be increased, and the more they are produced the more it will be increased (Cor. 12. 2.); and therefore they may be produced, till the distance between their extremities is equal to the distance between the parallels AB, CD (Cor. 12. 2.); and because parallel lines when produced, remain always at the same distance from each other (Cor. 11. 2.), BC will then meet AB. Therefore, if a straight line meet two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet upon that side on which are the angles, which are less than two right angles. Which was to be proved. PROP. XIV. THEOR. Parallelograms of equal base and altitude, are equal. Let the parallelograms ABCD, EBCF, be of equal base and altitude, they shall also be equal. Let the parallelograms be applied, so that the bases coincide with BC: and join AF. Then, because the altitude of the parallelogram ABCD, is equal to the altitude of the parallelogram EBCF, by the hypothesis, the perpendicular from F on BC, shall be equal to the perpendicular from A on BC (Def. 16. 2.); therefore the straight line AF shall be parallel to the straight line BC (10. 2.). But the straight line AD, passing through the point A, is also parallel to BC; and therefore AD coincides with AF: for the same reason EF coincides with AF. And because ABCD is a parallelogram, AD is equal to BC (9.2.). For the same reason, EF is equal to BC, and AD being equal to EF, and DE common, the whole or the remainder AE is equal to the whole or the remainder DF. AB is also equal to AC, and the two EB, AB, are equal to the two, FD, DC, each to each; and the angle FDC is equal to the angle EAB, because they are corresponding angles (3. 2.); therefore the triangle EAB is equivalent to the triangle FDC (4. 1.). From the quadrangle ABCF, take the triangle FDC, and from the same quadrangle take the triangle EAB, and the remainder, the parallelogram ABCD, will be equal to the remainder, the parallelogram EBCF. Q. E. D. COR. Equal parallelograms of equal base, are of equal altitude. PROP. XV. THEOR. A triangle is equal to half a parallelogram, of equal base and altitude. Let the triangle ECD, and the parallelogram ABDC, be of equal base and altitude: ECD shall be equal to half ABCD. Let them be applied so as to coincide with the base CD, and complete the parallelogram ECDF. Because the parallelograms ABDC and ECDF, are on the same base and of the same altitude, they are equal (14. 2.); but the triangle ECD is half the parallelogram ECDF; and therefore it is equal to half the parallelogram ABCD. Q. E. D. COR. I. Triangles of equal base and altitude, are equal. PROP. XVI. PROB. To describe a square upon a given straight line. Let AB be the given straight line; it is required to describe a square upon AB. E From the point A, draw AC at right angles (9. 1.) to AB, and make AD equal to AB; and through the point D, draw DE parallel (4. 2.) to AB, and through B draw BE parallel (4. 2.) to AD; therefore ABED is a parallelogram; whence AB is equal (9. 2.) to DE, and AD to BE; but AB is equal to AD; therefore the four straight lines AB, AD, BE, DE, are equal to one another, and the parallelogram D ABED is equilateral; likewise all its angles are right angles; because the straight line AD, meeting the parallels AB, DE, the angles BAD, ADE are equal to two right angles (1 Cor. 3. 2.); but BAD is a right angle, therefore also, ADE is a right angle; but the opposite angles of parallelograms are equal (9. 2.); therefore each of the opposite angles ABE, BED is a A B right angle; wherefore the figure ABED is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square (Def. 5. 2.), and it is described upon the given straight line AB. Which was to be done. COR. Hence every parallelogram that has one right angle, has all its angles right angles. PROP XVII. THEOR. In any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle, having the right angle BAC; the square described upon the side BC, is equal to the squares described upon AB, AC. the On BC describe (16. 2.) the square BDEC, and on AB, AC, squares BG, CH; and through A draw AL, parallel (4. 2.) to BD or CE, and join AD, CF; then, because each of the angles BAC, BAG, is a right angle (Def. 5. 2.), the two straight lines AC, AG, upon the |