Elementary Geometry

Miscellaneous Propositions

926 In a given line 𝐴𝐶, to find a point 𝑋 whose distance from a point 𝑃 shall have a given ratio to its distance in a given direction from a line 𝐴𝐵.
Through 𝑃 draw 𝐵𝑃𝐶 parallel to the given direction. Produce 𝐴𝑃, and make 𝐶𝐸 in the given ratio to 𝐶𝐵. Draw 𝑃𝑋 parallel to 𝐸𝐶, and 𝑋𝑌 to 𝐶𝐵. There are two solutions when 𝐶𝐸 cuts 𝐴𝑃 in two points.  Proof. By VI. 2 927 To find a point 𝑋 in 𝐴𝐶, whose distance 𝑋𝑌 from 𝐴𝐵 parallel to 𝐵𝐶 shall have a given ratio to its distance 𝑋𝑍 from 𝐵𝐶 parallel to 𝐴𝐷.
Draw 𝐴𝐸 parallel to 𝐵𝐶, and having to 𝐴𝐷 the given ratio. Join 𝐵𝐸 cutting 𝐴𝐶 in 𝑋, the point required. Proof. By VI. 2 928 To find a point 𝑋 on any line, straight or curved, whose distances 𝑋𝑌, 𝑋𝑍, in given directions from two given lines 𝐴𝑃, 𝐴𝐵, shall be in a given ratio.
Take 𝑃 any point in the first line. Draw 𝑃𝐵 parallel to the direction of 𝑋𝑌, and 𝐵𝐶 parallel to that of 𝑋𝑍, making 𝑃𝐵 have to 𝐵𝐶 the given ratio. Join 𝑃𝐶, cutting 𝐴𝐵 in 𝐷. Draw 𝐷𝐸 parallel to 𝐶𝐵. Then 𝐴𝐸 produced cuts the lines in 𝑋, the point required, and is the locus of such points. Proof. By VI. 2 929 To draw a line 𝑋𝑌 through a given point 𝑃 so that the segments 𝑋𝑃, 𝑃𝑌, intercepted by a given circle, shall be in a given ratio.
Divide the radius of the circle in that ratio, and, with the parts for sides, construct a triangle 𝑃𝐷𝐶 upon 𝑃𝐶 as base. Produce 𝐶𝐷 to cut the circle in 𝑋. Draw 𝑋𝑃𝑌 and 𝐶𝑌.
Then 𝑃𝐷+𝐷𝐶=radius therefore 𝑃𝐷=𝐷𝑋 But 𝐶𝑌=𝐶𝑋 therefore 𝑃𝐷 is parallel to 𝐶𝑌 (I 5, 28) therefore ⋯ by (VI. 2) 930 From a given point 𝑃 in the side of a triangle, to draw a line 𝑃𝑋 which shall divide the area of the triangle in a given ratio.
Divide 𝐵𝐶 in 𝐷 in the given ratio, and draw 𝐴𝑋 parallel to 𝑃𝐷. 𝑃𝑋 will be the line required.
𝐴𝐵𝐷: 𝐴𝐷𝐶= the given ratio (VI. 1), and 𝐴𝑃𝐷=𝑋𝑃𝐷 (I.37); therefore ⋯

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive