PROBLEM (k). To describe a triangle having two of its Ls equal to two given angles, and the side opposite to one of them equal to a given straight line. R A L. B A Let PAQ and B be the two given 4 s, and C the given straight line; produce PA to R and make 4 RAS = LB. = Hence QAS must be the remaining 4 of the required A. Take XY= C, and make s YXZ, XYZ= L s QAP, QAS. Then XYZ is the required ▲. N.B. This problem is impossible if the two given angles are not together less than two right angles. PROBLEM (2). To describe a triangle having two of its sides equal to two given straight lines, and the to a given L opposite to one of them equal Let A, B be the given straight lines and C the given ▲ . It is required to describe a ▲ having two of its sides equal to A, B and the opposite the latter = C. Make ▲ SRT = ▲ C, and cut off RS = A, and with centre S and radius = B describe a O. If this meet RT in T, join ST. Then the ▲ RST will be one which satisfies the conditions of the question. N.B. When C is either an obtusę z or a rt. the problem will admit of one solution if B be > A; otherwise the problem will be impossible. When C is an acute angle and B-the perpendicular from S on RT, then only one straight line can be drawn from SB, and there will thus be one solution. If B be<the perpendicular from S on RT, the problem is impossible. > If B be the perpendicular from S on RT, then two straight lines can be drawn from S to RT, each = B, which will be on the same side of SR if B be < A, and thus give two solutions; otherwise, if B be not < A, only one solution. THEOREM (m). If a straight line be divided into two equal parts and also two unequal parts, the squares on the unequal parts shall together be twice the squares on half the line and on the line between the points of section. Draw CF and AC; join AF. Then ACF is a right-angled isosceles ▲, and square on AF = twice the square on AC. (I. 35) Join FH; draw DK || to CF; KG || to AH; join AK. Then it may be easily shewn that FCH, KDH, FGK are right-angled isosceles a s, and that 4 AFH is a right ‹ . Hence square on FK = twice the square on GK, (1. 35) twice the square on CD; (1. 26) .*. squares on AF, FK = twice the squares on AC, CD; .. square on AK twice the squares on AC, CD. Also squares on AD, DH = squares on AD, DK, = = and ... = square on AK; (I. 35) twice the squares on AC, CD. COR. Let AH be bisected in C and produced to D: then shall the squares on AD, DH=twice the squares on AC, CD. This may be established in a similar manner to the above or deduced from it, thus: Produce DA and make AX = HD. Then XC = CD. .. squares on XH, HD = i.e. squares on AD, HD = twice the squares on XC, CH, twice the squares on DC, CA. THEOREM (2). The squares on the sides of a triangle are double the squares on half the base and on the line joining the vertex and the middle point of the base. From the vertex K of ▲ AKH draw KD 1 to AH. Δ Bisect AH in C, and join CK. Then shall the squares on AK, KH : on AC, CK. twice the squares For the squares on AD, DH= twice the squares on AC, CD; .. the squares on AD, DH and twice the square on DK: === twice the square on AC and twice the squares on CD, DK; i.e. the squares on AK, KH = twice the squares on AC, CK. Similarly, if the perpendicular from K fall on AH pro AC, CD, it follows that the squares on AK, KH = twice the squares on AC, CK. Euclid solves the following two problems, the necessity for which has been removed in this treatise by postulate 3. The corresponding postulate of Euclid is "Let it be granted that a circle may be described from any centre at any distance from that centre," which does not allow the transfer of distances by means of compasses. PROBLEM (p). From a given point to draw a straight line equal to a given straight line. Let C be the given point, and AB the given straight line. Join AC, and upon AC describe an equilateral a; with centre A and radius AB describe a cutting QA produced in R. B With centre Q and radius QR describe a cutting QC produced in X. Then CX shall be the line required. QA is = QC; ·.· QAC is an equilateral ▲ ; ... CX = AB; .. from C a straight line CX has been drawn |