. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. dius,describe a circle. Because TO is the bisectrix of the angle RTP, OR = OP(Prop. IV, Key); because CP is the bisectrix of the angle BCD,OR = OQ: hence, the circle OR passes through P and Q. Thiscircle is tangent to CT because CT is perpendicular to OR at R, andfor a like reason it is also tangent to CD at Q; the circle OR is tan-gent to the arc DPB, because CO = CP — OP: hence, the circleOR is the required circle. Prop. XLVL—Through a given point P,
. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. dius,describe a circle. Because TO is the bisectrix of the angle RTP, OR = OP(Prop. IV, Key); because CP is the bisectrix of the angle BCD,OR = OQ: hence, the circle OR passes through P and Q. Thiscircle is tangent to CT because CT is perpendicular to OR at R, andfor a like reason it is also tangent to CD at Q; the circle OR is tan-gent to the arc DPB, because CO = CP — OP: hence, the circleOR is the required circle. Prop. XLVL—Through a given point P, within a given angleABC, draio a circle that shall be tangent to both sides of that angle. Solution.—Let P be the given point and ABC the given angle. Draw the bisectrix BO, of thegiven angle, and also the line BP;from any point Q, of BO, draw QRperpendicular to AB, and from Q,as a centre, and with a radius QR,draw the arc RS, cutting BP in S;draw SQ and also draw PO parallel to SQ, intersecting BO at O; then draw OA parallel to QR, and withO as a centre and OA as a radius, describe the circle APC: also drawOC perpendicular to PROPOSITIONS FROM LEGEKDRE. 33 The triangles BSQ and BPO have the angle at B common, andsince QS is parallel to OP, the angles BQS and BOP are equal, as arealso the angles BSQ and BPO; hence, the triangles are similar(Bk. IV, Prop, 18), and consequently, their corresponding sides areproportional. We have, therefore, the proportion BQ : BO : : QS : OP . . (1). In like manner, we have from the triangles BQR and BOA, the proportion, BQ : BO : : QR : OA . . (2). Because the first three terms of proportions (1) and (2) are equal,each to each, their fourth terms must be equal, that is, OP = , the circle whose radius is OA, passes through P. The distanceOC is equal to OA (Prop. IV, Key); hence, the circle whose centreis O passes through C. Furthermore, BA is perpendicular to OA, atits extremity, and BC is perpendicular to OC, at its extremity;
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectgeometry, bookyear187
