Geometrical Problems Deducible from the First Six Books of Euclid, Arranged and Solved. to Which Is Added, an Appendix Containing the Elements of Plane Trigonometry
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1819 edition. Excerpt: ... join EF; and draw the diagonals AC, BD. The squares of AD, BC, AC, BD are equal to the squares of AB, DC, and four times the square of EF. Join AF, BF. Since AF bisects DC the base of the triangle ADC, AD + AC = 2DF + 2FA; and in the same manner, BC + BD = 2DF + 2FB-, (39.) If squares be described on the sides of a rightangled triangle; each of the lines joining the acute angles and the opposite angle of the square, will cut off from the triangle an obtuse-angled triangle, which will be equal to that cut off from the square by a line drawn from the intersection with the side to that angle of the square which is opposite to it. From the angles B, C of the right-angled triangle BAC, let lines BG, CD be D drawn to the angles of the squares described upon the sides, and from the intersections H and / let HE, IF be drawn to the opposite angles of the squares; the triangle BIC=AIF, and CHB = AHE. Join AG, AD. Then (Eucl. i. 37.) the triangle AFI= AIG; to each of which add ABI, .-. the triangle BIF= BAG--BCA (Eucl. i. 37.) From each of these equals take away the triangle BIA, and BIC= AIF. In the same manner it may be shewn that CHB = AHE. (40.) If squares be described on the two sides of a right-angled triangle; the lines joining each of the acute angles of the triangle and the opposite angle of the square will meet the perpendicular drawnfrom the right angle upon the hypothenuse, in the same point. Let BE, CF be squares described on the sides BA, AC containing the right angle. Join DC, BG; they intersect AL, which is perpendicular to BC, in the same point O. Produce BE, GFt to meet in H. Join HA, HB, HC. Let BH, CH respectively meet DC, BG in / and K. Since EH=AF=AC, and EA = AB, and the angles HE A, BAC are right angles, the triangles HE A, BAC are..
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- 04 Jul 2012
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