2. The difference BED is equal (1. 32.) to the two angles ECD, EDC, which (I. 5.) are equal : therefore ECD is half of BED. Again, in the triangles AHD, AHC, because AD, AC are equal, AH common, and the contained angles equal, the angles at H (I. 4.) are equal, and are therefore right angles. Then, in the triangles A EF, CEH, the angles AFE, CHE are equal, and AEC common; therefore (1. 32. cor. 5.) the angle EAF is equal to ECH, which has been proved to be equal to the half of BED. 3. Since the angle BAC is common to the triangles ABC, ADC, the angles ADC, ACD are (1.32.) equal to ABC, ACB; and there. fore, since ADC, ACD are equal, each of them is equal to half the sum of ABC, ACB; also either of them, ADC, is the complement of DAH, balf the vertical angle, since AHD is a right angle. 4. Because DH, HC are equal, and HE, DG parallel, GE is equal* (VI. 2.) to EC: and therefore BG, the difference of BE, GE, is also the difference of BE, EC. + PROP. VII. THEOR. In a triangle, straight lines drawn from the points of bisection of the three sides to the opposite angles, all pass through the same point. Let D, E, F be the points of bisection of the sides of the triangle ABC, and draw BE, CD, intersecting in G: bisect BG, CG in H, K, and join DE, HK. Then DE, HK are parallel and equal, each of them I being parallel to BC, and equal to half of it. Hence the triangles DGE, KGH are equiangular, and (I. 26.) HG is equal to GE, and KG to GD: and therefore BG is double of GE, and CG of GD. In the same manner, it would be proved, that a straight line joining AF would * It is easy to prove without proportion, that if AB (see the figure to the next proposition) be bisected in D, the straight line DE parallel to BC bisects AC, and that the triangle ADE is a fourth of ABC. For (1. 37.) the triangles BDC, BEC are equal. But (I. 38.) BDC is half of ABC; and there. fore BEC is half of ABC, and is equal to BEA. Hence the bases AE, EC are equal; for if they were not, the triangles ABE, CBE (I. 38. cor. 1.) would be unequal. Again, because AD, DB are equal, the triangle ADE is (I. 38.) half of ABE, and therefore a fourth of ABC. Conversely, if DE bisect AB, AC, it is parallel to BC. For (I. 38. cor. 2.) the triangles BDC, BEC are each half of ABC; and these being therefore equal, DE is parallel (I. 39.) to BC. Hence it is plain that the straight lincs joining DE, DF, EF, divide the triangle ABC into four equal triangles, similar to the whole and to one another. | Instead of cutting off AD equal to AC, the student may produce AC through C; and, cutting off, on AC thus produced, a part terminated at A, and equal to AB, and by making a construction similar to that of the foregoing proposition, he will find it easy to establish the same properties as those above demonstrated, or ones exactly analogous. # By the note to the preceding proposition ; or by VI. 2. and 4. R A D K B F divide BE into two segments, of which the one terminated in B would be double of the one terminated in E: that is, the straight line AF would pass through G. Schol. It appears from the demonstration, that each of the straight lines AF, BE, CD is divided at their common point into two segments, of which the one drawn to the angle of the triangle is double of the other. Cor. 1. Suppose HC were joined; then the three triangles CBH, CHG, CGE are (1.38.) all equal; and therefore BGC, being two-thirds of BEC, is one-third of BAC, which is double of BEC. Straight lines therefore drawn from G to the three angles, trisect the triangle ABC. Cor. 2. Hence the problem in which it is required to draw a straight line through a given point D, so that the part of it between two given straight lines AC, BC may be bisected in D, is solved simply by drawing DE parallel to BC, making EA equal to EC, and drawing ADB: or should the ratio of A D to DB not be that of equality, the solution will be the same, except that AE must have that ratio to EC. G G PROP. VIII. THEOR Four times the sum of the squares of the three straight lines drawn from the angles of a triangle to the points of bisection of the opposite sides, are equal to three times the sum of the squares of the sides. Let AF, BE, CD bisect the sides of the triangle ABC in the points F, E, D : four times the sum of the squares of AF, BE, CD, are equal to three times the sum of the squares of AB, BC, AC. For (II. A.) AB? + AC2 = 2BF9 + 2AF. From this, by doubling, and since (II. 4. cor. 2.) 4BF2 = BC2, there arises 2 AB2 + 2AC2 = BC2 + 4AF9. For the same reason, 2 AB + 2BCo = AC + 4BE”, and 2 AC: + 2BC = AB2 + 4CD!. Add these equals together, and from the sums take BC”, AC”, AB° : then 3AB2 + 3BC2 + 3AC2 = 4AF9 + 4BE2 + 4CD2. Cor. Since (App. I. 7. schol.) AF is treble of GF, the square of AF (I. 46. cor. 4.) is equal to nine times the square of GF; and therefore the squares of AF, BE, CD are evidently equal to nine times the squares of GF, GE, GD. Hence, by this proposition, thirty-six times the squares of GF, GE, GD are equal to three times the squares of AB, BC, AC; and consequently the squares of GF, GE, GD, are equal to one-twelfth of the squares of the B с sides AB, BC, CA. Hence also, since AG, BG, CG are double of GF, GE, GD, each of each, the squares of the former are quadruple of the squares of the latter ; and hence the sum of the squares of AG, BG, CG is one-third of the sum of the squares of the three sides AB, BC, AC. PROP. IX. THEOR, G F H LET ABC be any triangle, and on AB, AC, two of its sides, let any parallelograms ABMD, ACEF, be described : produce MD, EF to meet in G, and join GA; draw BH, CK parallel to AG, and meeting MG, EG, in H and K; join HK: HBCK is a parallelogram, and is equal to the two ABMD, ACEF. Produce GA to L. Then, since (const. and I. 34.) BH, CK are parallel and equal to AG, they are equal and parallel to one another: therefore also (I. 33.) HK is equal and parallel to BC, and (I. def. 24.) BCKH is a parallelogram ; and so also are LH, LK. Now (I. 35.) the parallelograms HL, HBAG are equal, because they are on the same base BH, and between the same parallels BH, LG: and the parallelograms HBAG, MBAD are equal, because they are on the same base AB, and between the parallels BA, MG. Hence (I. ax. 1.) MBAD is equal to Hl. In the same manner, it would be shown that ACEF is equal to LK: and therefore (I. ax. 2.) the whole parallelogram BK is equal to the two MBAD, ECAF. Schol. The 47th proposition of the first book is the case of this proposition in which BAC is a right angle, and the parallelograms AM, AE squares ; as in that case HBCK is also a square. For then DF is a parallelogram ; and in the triangles GFA, BAC, right-angled at F and A, FA is equal to AC, and GF to BA, each being equal to DA; therefore (I. 4.) BC is equal to AG, or to CK, which is equal to AG: and the angle ACB is equal to FAG, or to ECK, which (1. 34. cor. 4.) is equal to FAG. To the equal angles ACB, ECK, add KCÁ: then BCK is equal to ACE, which is a right angle. Hence the parallelogram HBCK B C is a square. PROP. X. THEOR. From B, C, the extremities of one of the sides of any triangle ABC, draw perpendiculars to the other sides, meeting those sides, or their continuations, in D and E; and on BC describe the square BG: BG is equal to the rectangles AB.BE, AC.CD; that is, to the rectangles contained by the sides and the segments intercepted on them, between the perpendiculars and the third side. H F A E Complete the parallelograms ABFH, ACGH; produce HF, and complete the rectangle AK. Then (I. 34. cor. 4.) the triangles ABC, HFG are equiangular, and (I. 26.) they are equal, since the sides BC, FG are equal. Take these triangles separately from the figure BCGHF, and there remains the square BG equal to the parallelograms AF, AG: and (I. 35.) the parallelograms AF, AK are equal. From the right angles KBE, FBC take FBA, and the remaining angles KBF, EBC are equal. Hence, since the angles at K and E are right angles, the triangles BFK, BCE are equiangular ; and the sides BF, BC are equal : hence (I. 26.) BK is equal to BE. Therefore the rectangle AK is contained by AB, BK, or by AB, BE; and it has been proved to be equal to AF. In the same manner, it would be shown that the rectangle AC.CD is equal to the parallelogram AG. Therefore the rectangles AB.BE and AC.CD are together equal to the parallelograms AF, AG; that is, as has been proved, to the square BG. Schol. If BAC be a right angle, the perpendiculars coincide with BA, CA, and the rectangles AB.BE, AC.CD become the squares of AB, AC: whence it appears that the 47th proposition of the first book, is also a case of this proposition. PROP. XI. THEOR. LET ABC be any triangle ; and from one of the angles let straight lines AD, AE be drawn to the opposite side, making with the other sides the angles BAD, CAE respectively equal to C, B, the remote angles at the base : the square of AB is equal to the rectangle CB.BD; and the square of AC to the rectangle BC.CE. For the triangles ABC, ABD have the angle B common and B DEC the angle C equal to BAD; therefore (VI. 4. cor.) BC : BA :: BA: BD. Hence (VI. 17.) the square of BA is equal to the rectangle BC.BD: and it would be proved in a similar manner, that the square of AC is equal to the rectangle BC.CE. Schol. If BAC be a right angle, the angles B, C being together equal to it, AD, AE will coincide, and the segments BD, EC will make up BC: and (II. 2.) the squares of AB, AC will be equal to the square of BC. This, therefore, is another proposition of which the 47th of the first book is a particular case. PROP. XII. THEOR. B IF through two given points two straight lines be drawn to meet in a given straight line, and to make equal angles with it, their sum is less than the sum of any other straight lines drawn through the given . points, and meeting in the given line. Let A, B be the given points, and CG the given line : draw ACE perpendicular to CG, making CE equal to CA, and draw EB cutting CG in F; join also AF: then (I. 4.) FA, FE are equal, and the angle AFC is equal to EFC, or (I. 15.) to BFG, so that AF, BF make equal angles with CG: AF, FB are together less than any other straight lines AD, DB drawn from A, and B to meet in CG. Join DE. Then (1. 4.) DE is equal to DA: add DB to each, and ED, DB are equal to AD, DB. But (1. 20.) ED, DB are greater than EFB, that is, than AF, FB, because EF, AF are equal: therefore also AD, DB are greater than AF, FB: and in the same manner it may be shown that AF, FB are less than two straight lines drawn to any other point in CG. If the given points be B and E, on the opposite sides of CG, it is plain that EF, FB are the shortest lines, and that they make equal angles with CG. E PROP. XIII. THEOR. The sum of two straight lines drawn from two given points, without a given circle, and meeting in the convex circumference, is a minimum, when they make, equal angles with the tangent at that point. Let A, B be two points without a given circle, and C a point |