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 A B 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 AC 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 в 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 G A B, 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 GA, the angle FCB to GA B, and the angle r 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 &; 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. PROP. VI. THEOR. 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 C A B and DA B, the sides DA and AB in the one, are equal to CB and BA с 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 вc 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 ACD, because the sides AC and AD are equal (Hyp.), the angles DCE and C D F 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 demon. strated. C D A A B If two triangles have two sides of the one respectively equal to two sides of the other, and also have their bases equal, the angles opposite the equal bases are equal, and the two triangle. are equal in all respects. Let ABC and DEF be two triangles, having the side AB equal to DE, the side B C to EF, and the base AC equal to the base DF; then the angle в is equal to the angle E; and the two triangles are equal in all respects. B F For if the triangle ABC be so applied to the triangle D E F, that the point A may be on D, and the base A c upon D F, then the point c must coincide with F, because AC and DF are equal (Hyp.); and as the base AC coincides with DF, and the sides A B and BC are equal to the sides DE and EF (Hyp.), the vertex B must fall on E (Prop. 7.), and the equal sides must coincide (Ax. 10.); therefore the angles B and E must also coincide, and are therefore equal (Ax. 8.). And as the three sides of one triangle coincide with the three sides of the other, the triangles themselves must also coincide, and are therefore equal (Ax. 8.). Which was to be demonstrated. COR. Two triangles are equal, if they have a common side, and the remaining sides of the one equal to the remaining sides of the other. PROP. IX. PROB. To bisect a given rectilineal angle. Let BAC be the given angle, it is required to bisect it. E Take any point D in the side A B, and from AC cut off a part A E equal to A D (Prop. 3.) : join DE, and upon it describe an equilateral triangle DEF (Prop. 1.) at the side opposite to A; join A F; then the angle BAC or DAE is bisected by the line a F. For in the triangles DAF and EAF, the side DA is equal to E A (Const.), and a F is common to both, and the base DF is equal to E F (Const.); therefore the angle DAF is equal to EAF (Prop. 8.), and therefore the given angle BAC is bisected by the line A F. Which was to be done. B F C PROP. X. PROB. To bisect a given right line. Let A B be the given line, it is required to bisect it. Upon A B describe an equilateral triangle ABC (Prop. I.), and bisect the angle ACB by the line CD (Prop. 9.); then AB is bisected in the point D. A D B For in the triangles ACD and BCD, the side A c is equal to BC (Const.), the side CD is common to both, and the angle A CD is equal to BCD (Const.); therefore the base AD is equal to BD (Prop. 4.), and therefore the given line AB is bisected in D. Which was to be done. COR. By this proposition a line may also be divided into 4 equal parts, or into 8, 16, &c., by bisecting again each part. PROP. XI. PROB. To draw a right line perpendicular to a given line from a given point in it. Let A B be the given line, and c the given point; it is required to draw from c a line which shall be perpendicular to the line AB. G B F Take any point D in AC, and cut off from CB a part cr equal to CD (Prop. 3.), A and upon DF describe an equilateral triangle DFG (Prop. 1.); join &c; then the line GC is perpen dicular to A B. D C For in the triangles CDG and CFG, the sides CD and DG in the one, are equal to the sides CF and FG in the other (Const.), and the side CG is common to both, therefore the angles DCG and FCG opposite the equal sides are equal (Prop. 8.), and these are the adjacent angles; therefore the line Gc is perpendicular to A B (Def. 11.), and it is drawn from the point c. Which was to be done. COR. In like manner a perpendicular can be drawn from the extremity of a given line, by first producing the line. B |