To describe an equilateral triangle upon a given right line. Let AB be the given line; it is required to describe an equilateral triangle upon it. E B From the centre A, with the radius A B, describe the circle BCD (by Post. 3.), and from the centre B, with the radius BA, describe the circle ACE, cutting the former in c; join CA and CB (Post. 1.); then the triangle ABC is equilateral. For the line AC is equal to A B, being radii of the same circle BCD (Def. 16.); and the line BC is equal to BA, being radii of the same circle ACE: since then each of the lines AC and BC is equal to the same line AB, they are equal to one another (Ax. 1.); therefore the three lines A B, BC, and CA are all equal; therefore the triangle ABC is equilateral; and it is described upon the given line AB. Which was to be done. SCHOL Another equilateral triangle may be obtained by drawing right lines from A and B to the point G, in which the circles cut at the other side of the line AB, which can be demonstrated in the same manner. PROF. II. PROB. From a given point, to draw a line equal to a given right line. Let A be the given point, and BC the given line; it is required to draw from ▲ a line equal to BC. H F G A D B E Join A and B (Post. 1.), and upon the line AB describe an equilateral triangle DAB (Prop. 1.), and from the centre в with the radius BC, describe the circle CEF (Post. 3.), and produce the line DB till it meets the circumference in E (Post. 2.); then from the centre D, with the radius DE, describe the circle EGH (Post. 3.), and produce the line DA till it meets the circumference in G: then the line AG is equal to the given line B c. For the line DG is equal to DE, being radii of the same circle EGH (Def. 16.), and the part DA is equal to DB, being sides of an equilateral triangle (Const.), therefore the remaining part AG is equal to BE (AX. 3.); but the line BC is also equal to BE, being radii of the same circle CEF: since then each of the lines AG and BC is equal to the same line BE, they are equal to one another(Ax. 1.); therefore the line AG is equal to the given line BC, and it is drawn from the given point a. Which was to be done. PROP. III. PROB. From the greater of two lines, to cut off a part equal to the less. Let AB be the greater of two lines, and CD the less; it is required to cut off from AB, a part equal to CD. From the point A draw a line AG equal to CD (Prop. 2.), and from the centre A, with the radius AG, describe a circle (Post. 3.) cutting the line AB in E; then AE is equal to CD. B E For the line AE is equal to AG, being radii of the same circle (Def. 16.); but the line CD is also equal to AG (Const.): since then each of the lines AE and CD is equal to the same line AG, they are equal to one another (Ax. 1.); therefore AE is equal to CD, and it is cut off from the greater line a B. Which was to be done. COR. By a similar construction the following problem may be solved: To produce the less of two given lines, till it be equal to the greater. PROP. IV. THEOR. If two triangles have two sides and the contained angle in the one, respectively equal to two sides and the contained angle in the other, their bases or third sides are likewise equal, and the remaining angles of the one, are respectively equal to the remaining angles of the other, and the two triangles are equal in all respects. Let ABC and DEF be two triangles, having the side AB equal to DE, the side AC equal to DF, and the angle a equal to the angle D; then the base B C is equal to EF, the angle в equal to the angle E, and the angle c is equal to the angle F; and the two triangles are equal in all respects. Α A B D A C E F For, if the triangle ABC be so applied to the triangle DEF, that the point A may be on D, and the side AB upon DE, then the point в must coincide with E, because the sides A B and DE are equal (Hyp.); and the side A C will fall on DF, because the angles A and D are equal (Hyp.); and the point c must coincide with F, because the sides Ac and DF are equal (Hyp.): and as the points B and c coincide with the points E and F, the lines BC and EF must also coincide, if not two right lines would inclose a space, which is impossible (Ax. 10.). Therefore the base BC coincides with EF, and is equal to it (Ax. 8.). And as the sides of the angle B, coincide with the sides of the angle E, the angles themselves must coincide, and are therefore equal (Ax. 8); also the angles c and F must coincide, and are therefore equal. And as the three sides of one triangle coincide with the three sides of the other, the triangles themselves must coincide, and are therefore equal in all respects (Ax. 8.). Which was to be demonstrated. COR. 1. Two triangles are equal, which have an angle common to both, and the two sides which contain it in the one, equal to the two sides which contain it in the other. COR. 2. Two triangles are equal, which have a side common to both, and another side with the contained angle in the one, equal to another side with the contained angle in the other. PROP. V. THEOR. The angles at the base of an isosceles triangle are equal, and if the equal sides be produced, the angles below the base are also equal. Let ABC be an isosceles triangle, having the sides AB and BC equal; then the angles BAC and BCA at the base are equal, and if the equal sides be produced, the angles FAC and GCA below the base are also equal. Take any point F, in one of the sides produced, and cut off from the other a part BG equal to BF (Prop. 3.): join AG and FC. Then in the triangles FCB and GAB, the sides FB and BC in the one, are equal to the sides G B F A B and BA in the other (Hyp.), and the angle B is common to both; therefore the two triangles are equal in all respects (Prop. 4.), the base Fc is equal to G A, the angle FCB to GAB, and the angle F equal to the angle G. And if from the equal sides BF and BG, there be taken the equals BA and BC, the remainders A F and CG are also equal (Ax. 3.); then in the triangles AFC and CGA, the sides AF and FC in the one, are equal to the sides CG and GA in the other, and the angle r is equal to the angle G; therefore the two triangles are equal in all respects (Prop. 4.), the angles FAC and GCA are equal, and these are the angles below the base; also the angles FCA and GAC are equal; and if from the equal angles FCB and GAB, there be taken the equals FCA and GAC, the remainders BAC and BCA are also equal (Ax. 3.), and these are the angles at the base. Which was to be demonstrated. If two angles of a triangle be equal, the sides opposite to them are also equal. In the triangle CAB, let the angles C A B and CBA be equal; the sides CA and C B are also equal. For if they be not equal, let one of them ca be greater than the other, and from it cut off DA equal to B C (Prop. 3.); and join B D. Then in the triangles CAB and DA B, the sides DA and AB in the one, are equal to CB and BA D c B in the other (Hyp.), and the angles CAB and C BA are equal (Hyp.), therefore the triangles are equal (Prop. 4.); a part equal to the whole, which is absurd; therefore neither of the sides CA and CB is greater than the other: they are therefore equal. Which was to be demonstrated. PROP. VII. THEOR. If two triangles, on the same base, and on the same side of it, have the sides which terminate in one end of the base equal, those sides which terminate in the other end of the base are not equal. Let the triangles ACB and ADB, on the same base AB, and on the same side of it, have the sides AC and AD equal; then the sides BC and BD are not equal. CASE 1. When the vertex of each triangle is without the other. A D B Join CD, and if it be possible, let Bc be equal to BD then in the triangle CAD, because the sides AC and AD are equal (Hyp.), the angles ACD and ADC are also equal (Prop. 5.); but the angle ACD is greater than BCD (Ax. 9.); therefore the angle ADC is greater than BCD; much more than is the angle BDC greater than BCD. But in the triangle B CD, the sides BC and BD are equal (Hyp.); therefore the angles BCD and BDC are equal (Prop. 5.); but the angle BDC has been proved to be greater than B CD; greater than, and equal to it, which is absurd; therefore the sides BC and BD are not equal. CASE 2. When the vertex D of one triangle is within the other. Produce the sides AC and AD to E and F: then in the triangle AC D, because the sides AC and AD are equal (Hyp.), the angles DCE and CDF are also equal (Prop. 5.); but the angle DCE is greater than DCB (Ax. 9.), therefore the angle CDF is greater than DCB; much more than is the angle CDB greater than DC B. But in the A E B triangle B C D, the sides BC and BD are equal (Hyp.), therefore thè angles BDC and BCD are equal (Prop. 5.): but the angle BDC has been proved to be greater than B CD; greater than, and equal to it, which is absurd; therefore the sides BC and BD are not equal. CASE 3. When the vertex D of one triangle falls on a side of the other B C. Then it is evident that the whole line BC is greater than its part BD (Ax. 9.). Therefore in no case can the side BD be equal to BC. Which was to be demonstrated. A B |