PROP. XLVI. PROB. GEN. ENUN.-To describe a square upon a given straight line. PART. ENUN.-Let AB be the gn. st. line; then it is required to describe a square upon it. CONST. and DEMONST.1. From the pt. a draw ac at rt. s to AB (Prop. XI.); make AD = AB (Prop. III.); C D through D draw DE || to AB, and through B draw BE to AD (Prop. XXXI.); .. ABED is am; and ... AB = DE, and AD = BE. (Prop. XXXIV.) 2. But AB AD (Const.); .. AB = AD = 2 rt. Zs ms 0 (Const.); are 4. Now the opposites of (Prop. XXXIV.); ... each of the opposite SABE, BED is a rt. . 5. Hence the figure ABED is equiangular as well as equilateral; it is .. a square (Def. 30), and it has been described upon the gn. line AB.-Q. E. F. COR.-Hence every m that has one rt. has all its 4s rt. Zs. Although not immediately deducible from the above Proposition, the following Theorem may be inserted here. PROP. BBB. THEOR. GEN. ENUN.-If in each of the sides of a square, a point be taken at equal distances from the four angles, the straight lines which join them will also form a square. PART. ENUN.-Let ABCD be a square. = points at distances from the four 2. Join EF, FG, GH, HE. Then the figure EFGH is also a square. DEMONST.-1. Since AE = BF, and AHEB (Const.); ... iu the A AEH, BEF, the two EA, AH = FB, BE, each to each, and the rt. / EAH = rt. EBF; ... the base EH = base EF, and the BEF = ▲ AHE. (Prop. IV.) 2. In like manner, FG and GH = 3. Hence EF = FG = GH = HE figure EFGH is equilateral. E, F, G, H, (Ax. 1), and .. the 4. Again, the exterior / HEB of ▲ AEH = interior s EAH AHE (Prop. XXXII.), and the ▲ BEF = ▲ AHE; .., by subtraction, / HEF = LEAH (Ax. 3) = art. 4. (Hyp.) 5. In the same manner the 4s at F, G, H may be shewn to be rt. s. 6. Hence the figure EFGH is also equiangular, and is a square (Def. 29). Wherefore, if in each of the sides &c.-Q. E. D. ... it GEN. ENUN. In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. PART. ENUN.-Let ABC be a rt. d, GB, HC (Prop. XLVI.); D K through a draw AL | to BD or CE (Prop. XXXI.), and join AD, FC (Post. 1). S DEMONST.-1. Now, because each of the BAC, BAG is a rt. (Const.) ... at the pt. a in the st. line AB, the line CA, AG, on opposite side of it, makes the adjacent s = to two rt. ≤s; and ... CA is in the same st. line with AG. (Prop. XIV.) 2. For the same reason AB and AH are in the same st. line. 3. Again, because the rt. DBC = rt. ≤ FBA (Ax. 11), add to each the ▲ ABC; ... the DBA FBC. 4. Hence, in the AS ABD, FBC, the two AB, BD = FB, BC, each to each (Const.), and the included ABD included FBC; ... the base AD = base FC, and the ▲ ABD = Δ FBC. (Prop. IV.) 5. But the □m BL = 2 ▲ ABD, upon the same base BD, and between same || AL, BD; S and the square GB = 2 FBC, upon the same base FB, and between same || FB, GC. (Prop. XLI.) 6. Now the doubles of = one another (Ax. 6); .`. the GB. things are to m BL = square 7. In the same manner, it may be proved, by joining AE, KB, that the □m CL = square HC. 8. Hence, by addition, the square upon BC = sum of the squares upon AB and AC, or вC2 = AB2 + AC2. Wherefore, in any rt. d -Q. E. D. Zda, the square &c. This Proposition is no less remarkable for its elegance than for its extensive utility. A few deductions are added for the student's consideration, who will observe, that the side which subtends the rt. 4 in a rt. ▲d ▲A, is called the hypothenuse, from the verb vπоTεívεiv, to subtend. PROP. CCC. THEOR. GEN. ENUN. The square described upon the diameter of a rectangular parallelogram, is equal to the square described upon two of its unequal sides. PART. ENUN.-Let ABCD be a A rectangularm, whose diam. is BD; then BD2 DC2 + BC2. == = AB2 + AD2 DC2 + BC2. (Prop. XLVII.) Wherefore the square &c. B D COR.-If the m be a square, the square upon the diameter is equal to twice the square upon either of the sides; for in this case AB = 2 AB2. AD; .. BD2 AB2 + AD2 = PROP. DDD. THEOR. GEN. ENUN.-If from one of the acute angles of a rightangled triangle, a line be drawn to the opposite side, the squares of that side and the line so drawn are, together, equal to the squares of the segment adjacent to the right angle and the hypothenuse. PART. ENUN.-Let ABC be a rt. L ABC. From A draw AD to any pt. in the opposite side BC; then the squares upon AD and BC = the squares upon AC and BD. DEMONST.-For since Ac2 =3 AB2 + BC2 (Prop. XLVII.); add BD2 to each side; AC2 + BD2 AB2 + = AD2 + BC2. B Wherefore, if from one &c. Q. E. D. d▲, having the rt. A PROP. EEE. THEOR. GEN. ENUN.-In any triangle, if a line be drawn from the vertex perpendicular to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base. PART. ENUN.-Let ABC be any A; from the vertex A draw AD to the base BC; then the difference of the squares upon AB and AC = difference of squares upon BD and DC. DEMONST.-Because ADC is a rt. 2d A, .. Ac2: AD2 + DC2; and because ABD is a rt. 4d A, .. AB2 = AD2 + BD2; .., by subtraction, в D AC2 AB2 = DC2 BD2. Wherefore in any A, if a line &c.-Q. E. D. L |