at the base + the vertical = two right; if ... the vertical ▲ be a right, and if it be taken from both, there remains the double of the ▲ at the base a right 4;.. the at the base half of a right . But, if the vertical be obtuse, when it is taken away from the same equals as before, there will remain the double of the at the base equal to a less than a right ▲; ... the at the base is, in this case, less than the half of a right . And, in like manner it may be shewn, that, when the vertical of the isosceles A is acute, the at the base is greater than the half of a right angle. PROP. XXVIII. 38. THEOREM. If either of the equal sides of an isosceles triangle be produced, towards the vertex, the straight line, which bisects the exterior angle, shall be parallel to the base. at For, (E. 5. 1. and E. 32. 1.) the exterior the vertex of an isosceles A is the double of either of the at the base; ... the half of that interior is equal to either of the .. the straight line bisecting the (E. 28. 1.) parallel to the base. at the base; vertical is PROP. XXIX. 39. THEOREM. The distance of the vertex of a triangle from the bisection of its base, is equal to, greater than, or less than the half of the base, accordingly as the vertical angle is a right, an acute, or an obtuse angle. First, let ABC be a right-angled A, right angled at A, and let AD join A and the bisection, D, of the base: AD=DB, or DC. For, if not, AD is either greater or less than BD: Produce BA to X; and first, if it be possible, let AD > DB; .., also, AD > DC; .. (E. 18. 1.) <B> <BAD, and <C> < CAD; .. 2 B+2 C> < BAD+4 CAD; i. e. 2 B+ <C> < BAC; but (hyp. and E. def. 10. 1.) <BAC=CAX; .. < B+<C> <CAX; which (E. 32. 1.) is absurd. And, in like manner, if DA be supposed to be less than BD, it may be shewn that 2 B + C < CAX; which is absurd. Therefore, DA=DB, or DC. Next, let the vertical / CEB, of the A EBC, be acute, and let ED join E and the bisection, D, of BC, ED> BD, or DC. From either of the B or C, as C, if the A EBC be acute-angled, draw (E. 12. 1.) CAL to the opposite side EB; and join A, D: Then (S. 7. 1.) CA falls within EB; and, since (constr.) the CAE is a right, the DAE is greater than a right; .. (E. 17. 1.) the AED is less than a right, and .. less than the DAE; .. (E. 19. 1.) DE>DA; but, by the former case, DA=DB; ... DE>DB, or DC. Lastly, if FBC be an obtuse-angled A, obtuseangled at F, join F, D; draw, as before, CA BF; and join A, D: Then (S. 7. 1.) CA falls without BF, and the AFD (E. 16. 1.) > the ▲ FBD; but since (1st case) DA DB, the ▲ DBF = L DAF (E. 5. 1.); .. ¿ AFD > ▲ DAF; .. (E. 19. 1.) DA> DF; but DA=DB; .. DF< DB, or DC. Or, the two last cases may be proved, ex absurdo, in the same manner as the first is proved. 40. COR. 1. If any number of triangles have a right angle for their common vertical angle, and have equal hypotenuses, the locus of the bisections of the several hypotenuses is a quadrantal arch of a circle, having the common vertex for its centre, and the half of any hypotenuse for its radius. For, the bisections of the hypotenuses will, each of them, (S. 29. 1.) be at a distance from the D 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 ▲ DAB= 2 DBA: also, since G is the bisection of BC, .. (S. 29. 1.) GA=GB; .. (E. 5. 1.) the GAB=2 GBA; ... the GAB/DAB, the greater to the less, which is absurd; .. no other point than D can be the bisection of BC. L 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 1 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. |