4. The difference of the sides of a triangle is less than the difference of the segments of the base, made by the perpendicular from the vertex. 5. The difference between the square on one of the equal sides of an isosceles triangle, and the square on any line drawn from the vertex to a point in the base, is equal to the rectangle contained by the segments of the base. 6. The square on either side of a right-angled triangle is equal to the rectangle contained by the sum and the difference of the hypotenuse and the other side. PROP. VI.—Theorem. If a line (AB) be bisected (at C), and divided externally in any point (D), the rectangle (AD.BD) contained by the segments made by the external point, together with the square on half the line, is equal to the square on the segment between the middle point and the point of external division. Dem.—On CD describe the square CDFE [I. xlvi.], and join DE; through B draw BHG parallel to CE [I. xxxi.], meeting DE in H; through H draw KLM parallel to AD; and through A draw AK parallel to CL. Then because AC is equal to CB, the rectangle AL is equal to CH [I. xxxvi.]; but the complements CH, HF are equal [I. xliii.]; therefore AL is equal to HF. To each of these equals add CM and LG, and we get AM and LG equal to the square CDFE; but AM is equal to the rectangle AD.DM, and therefore equal to the rectangle AD.DB, since DB is equal to DM; also LG is equal to the square on CB, and CDFE is the square on CD. Hence the rectangle AD.DB, together with the square on CB, is equal to the square on CD. Or thus:— Dem.—On CB describe the square CBEF [I. xlvi.]. Join BF. Through D draw DG parallel to CF, meeting FB produced in H. Through H draw KM parallel to AB. Through A draw AK parallel to CL [I. xxxi.]. The parallelogram CM is equal to DE [I. xliii.]; but AL is equal to CM [I. xxxvi.], because they are on equal bases AC, CB, and between the same parallels; therefore AL is equal to DE. To each add CH, and we get the parallelogram AH equal to the gnomon CMG; but AH is equal to the rectangle AD.DH, and therefore equal to the rectangle AD.DB, since DH is equal to DB [iv., Cor. 1]; therefore the rectangle AD.DB is equal to the gnomon CMG, and the square on CB is the figure CE. Therefore the rectangle AD.DB, together with the square on CB, is equal to the whole figure LHGF—that is, equal to the square on LH or to the square on CD. Or thus:AD = AC + CD = CD + CB; BD = CD − CB. Hence AD.DB = (CD + CB)(CD − CB) = CD2 − CB2; therefore AD .DB + CB2 = CD2. Exercises. 1. Show that Proposition vi. is reduced to Proposition v. by producing the line in the opposite direction. 2. Divide a given line externally, so that the rectangle contained by its segments may be equal to the square on a given line. 3. Given the difference of two lines and the rectangle contained by them; find the lines.