. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. P, cutting DE in Q,and prolong it. The line DE is parallel to BC; conse-quently the triangles PFB and PQE are similar, as are also the trian-gles PFC and PQD. From PFC and PQD, we have, PF : PQ : : FC : QD . . (1).From the triangles PFB and PQE, we have, PF : PQ : : FB : QE . . (2). From (1) and (2), we have, FC : QD : : FB : QE. But, FC = FB, hence, QD = QE; the line FP must thereforepass through the middle of DE, and therefore, from Prop. LVI, Key,it
. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. P, cutting DE in Q,and prolong it. The line DE is parallel to BC; conse-quently the triangles PFB and PQE are similar, as are also the trian-gles PFC and PQD. From PFC and PQD, we have, PF : PQ : : FC : QD . . (1).From the triangles PFB and PQE, we have, PF : PQ : : FB : QE . . (2). From (1) and (2), we have, FC : QD : : FB : QE. But, FC = FB, hence, QD = QE; the line FP must thereforepass through the middle of DE, and therefore, from Prop. LVI, Key,it must also pass through A, that is, the line through A and P is thethird median of the given triangle. Hence, the three medians passthrough P, which was to be proved. Prop. LVIII.— On the sides AB and AC of any triangle ABC,construct any two parallelograms ABDE and ACFG; prolong thesides DE and FG till they meet in H; draw HA, and on the third sideof the triangle BC, construct a parallelogram two of whose sides areparallel and equal to HA: then show that the parallelogram on BC isequal to the sum of the parallelograms on AB and 42 KEY. Demonstration.—Let ABC be any triangle; Let AD and AF beany parallelograms constructed on AB andAC as sides; and let H be the point inwhich DE and FG meet, when prolonged;let CP be a parallelogram whose sides BPand CQ are parallel and equal to HA. Draw HA, and prolong it to S; alsoprolong PB and QC, to K and L. The parallelograms ABDE and ABKHhave a common base AB, and a commonaltitude; hence, they are equal (Bk. IV, Prop. 1); the parallelograms ABKH and SPBT have equal bases,HA and ST, and a common altitude; they are therefore equal: hence,the parallelogram ABDE is equal to the parallelogram STBP. Inlike manner it may be shown that the parallelogram ACFG is equalto the parallelogram STCQ. The sum of the parallelograms TP andTQ is equal to the parallelogram BCQP; hence, the parallelogramBCQP is equal to the sum of the parallelograms ABDE
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Keywords: ., bookcentury1800, bookdecade1870, booksubjectgeometry, bookyear187
