common vertex equal to the half of one of the equal hypotenuses; i. e. they will all be at distances from that point, equal to the half of any one of those equal lines: It is manifest, ..., that they will be in the circumference of a circle, described from that point as the centre, at a distance equal to the half of one of the hypotenuses. 41. COR.2. A circle described from the bisection of the hypotenuse of a right-angled triangle as a centre, at the distance of half the hypotenuse, will pass through the summit of the right angle. 42. COR. 3. The vertical angle of a A being a right angle, a point in the base, which is equidistant from the vertex and from either extremity of the base, bisects the base. Let the point D, in the base BC of the ABC, having the ∠ B a right angle, be equidistant from either extremity, as B, of BC, and from the angular point A: The point D bisects BC.. For, if not, let G be the bisection of BC, and join D, A and E, A: Then, since (hyp.) DA= DB, ... (E. 5. 1.) the 2 DAB= ∠ DBA: also, since G is the bisection of BC, ... (S. 29. 1.) GA=GB; .. (E. 5. 1.) the GAB=∠GBA; ... the GAB=∠DAB, the greater to the less, which is absurd; ... no other point than D can be the bisection of BC. PROP. XXX. 43. PROBLEM. Upon a given finite straight line, as a diameter, to describe a square. Let AB be a given finite straight line: Upon AB, as a diameter, it is required to describe a square. Bisect (E. 10. 1.) AB in E; through E draw (E. 11. 1.) DEC to AB, and make (E. 3. 1.) ED and EC each of them equal to AE or EB: Join A, D, and D, B, and B, C, and C, A: The figure ADBC is a square, having AB for its di ameter. For since (constr.) DE=EC, and AE is common to the AED, AEC, and that the right ∠ AED = right ∠ AEC, .. (E. 4. 1.) AD=AC; and in the same manner AD may be shewn to be equal to DB, and DB to BC; ... the figure ADBC is equilateral. Again, since (constr.) AE=DE, the ∠ EAD =∠EDA (E. 5. 1.); but (constr.) AED is a right ; .. each of the EAD, EDA, is half a right ; and, in the same manner, may each of the EDB, DBE, CBE, BCE, ECA, EAC, be shewn to be half a right ; ... all the 4 of the figure ADBC are right; and it has been proved that all its sides are equal; .. (E. 30. def. 1.) ADBC is a square. PROP. XXXI. 44. THEOREM. If either of the acute angles of a given right-angled triangle be divided into any number of equal angles, then, of the segments of the base, subtending those equal angles, the nearest to the right angle is the least; and, of the rest, that which is nearer to the right angle is less than that which is more remote. Let ACB be a right-angled A, right-angled at C, and let the acute ∠ BAC be divided into any number of equal, CAD, DAE, EAB, &c.; then is CD the least of the segments of the base subtending those equal, and of the rest DE<EB; and so on. For, at the point D in AD make (E. 23. 1.) the ∠ ADF=∠ ADC: And since, also, the ∠ CAD = ∠ DAE (hyp.) and AD common to the two ACD, AFD, ... (E. 26. 1.) DF=DC: But (E. 19. 1. and E. 32. 1.) DE> DF; ... DE > DC; i. e. DC < DE. Again, at the point E, in AE, make the ∠ AEK =∠AED; and it may, in like manner, be shewn that EK=ED: But (E. 16. 1.) ∠ BKE > ∠ AEK; ∴∠BKE > < AED; and ∠ AED > ∠ ABE; much more then is ∠ BKE > < EBK; .. (E. 19. 1.) ВЕ > EK or ED; i. e. ED < EB. And in the same manner may EB be shewn to be less than the next segment that is more remote from C; and so on. 45. Cor. It is manifest, from the demonstration, that if any three straight lines AB, AE, AD, be drawn to the given straight line XC from a given point A, without it, so that the 2 BAE= <EAD, the segment BE, of XC, which is the further from the perpendicular AC, shall be greater than the segment ED, which is the nearer to AC. PROP. XXXΧΙΙ. 46. THEOREM. If either angle at the base of a 1 triangle be a right angle, and if the base be divided into any number of equal parts, that which is adjacent to the right angle shall subtend the greatest angle at the vertex; and, of the rest, that which is nearer to the right angle shall subtend, at the vertex, a greater angle than that which is more remote. Let ACB be a right-angled A, right-angled at C, and let the base BC be divided into any number of equal parts CD, DH, HG, &c. Of these segments DC shall subtend the greatest at the vertex A; and of the rest DH shall subtend, at A, a greater & than HG; and so on. For, join A, D, and A, H, and A, G, &c.; also, at the point A, in DA, make (E. 23. 1.) the DAE=CAD: Then (S. 31. 1.) ED > DC; but (hyp.) DC=DH; .. ED>HD, and it is manifest that the EAD > ∠ HAD; but (constr.) ∠EAD=∠CAD; .. CAD > ∠DAH : And, in the same manner, it may be shewn, by |