N.B.—It will assist the learner to note the following steps in this Proposition : a. To prove that AEB, i.e. AEG, is a right angle (as in 1). b. To prove, in the triangle BDG, that BD = DG (as in 2). c. To prove, in the triangle EGF, that FE = FG (as in 3). d. To prove, also in the triangle EGF, that the square on EG = double the square on EF, i.e. on CD (as in 4). e. To prove in the triangle ACE that the square on AE = double the square on AC (as in 5). f. To prove from the triangles AEG and ADG, in connection with previous results, the assertion made in the Proposition. Exercise. Prove the other case in Prop. X. that if BA be produced to D, beyond A, The squares on BD and AD = double the squares on BC and CD. F S PROP. XI. PROBLEM. To divide a straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part. Let AB be a straight line. It is required to divide AB into two parts, say in H, so that CONSTRUCTION.-1. On AB describe the square ACDB (I. 46). 2. Bisect AC in E and join BE (I. 10, and post. 1). 3. Produce CA to F, making EF EB (post. 2, and I. 3). 4. On AF describe the square AFGH. 5. Produce GH to K, in CD. Then it is to be proved that AB is divided in H, so that the square on PROOF.—The rect. CF, FA,) = { (II. 6). with the square on EA EF the square on EB, since EF EB (cons.). the squares on EA and AB (I. 47), the squares on EA and AB; = the square on AB (ax. 3). But FK is the rect. CF, FA, since FG = FA and AD is the square on AB (cons.), therefore FK = AD. Take away the common part) the remainder HD. Now, FH is the square on AH (cons.), and HD is the rectangle DB, BH, i.e. AB, BH (cons.). Therefore, it is proved, as required, that The rectangle AB, BH = the square on AH; and the straight line AB is divided into two parts in H, as required. Q. E. F. PROP. XII. THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and, from the acute angle A, let AD be drawn perpendicular to BC, produced to D. Then it is to be proved that The square on AB is greater than the squares on BC and CA by twice the rect. BC, CD. = PROOF.-Because the square on BD the squares on BC and CD, and twice the rect. BC, CD (II. 4), therefore the squares on BD and DA the squares on BC, CD, and DA, and twice the rect. BC, CD (ax. 2). = But because BDA is a right angle (hyp.), therefore the square on AB = the squares on BD and DA (I. 47). Similarly, the square on CA = the squares on CD and DA. Therefore, substituting these values (in line 2), the square on AB= the squares on BC and CA, with twice the rect. BC, CD. Therefore, it is proved, as required, that The square on AB is greater than the squares on BC and CA by twice the rectangle BC, CD. Wherefore, In obtuse-angled triangles, &c. Q. E. D. Exercise. Prove the other case in Prop. XII. that if the perpendicular BD be drawn from the other acute angle, B, to the side AC produced to D, The square on AB is greater than the squares on BC and CA by twice the rectangle AC, CD. |