Elementary Geometry

Miscellaneous Propositions

931 To divide the triangle 𝐴𝐵𝐶 in a given ratio by a line 𝑋𝑌 drawn parallel to any given line 𝐴𝐸.
Make 𝐵𝐷 to 𝐷𝐶 in the given ratio. Then make 𝐵𝑌 a mean proportional to 𝐵𝐸 and 𝐵𝐷, and draw 𝑌𝑋 parallel to 𝐸𝐴.
Proof: 𝐴𝐷 divides 𝐴𝐵𝐶 in the given ratio (VI. 1). Now 𝐴𝐵𝐸∶𝑋𝐵𝑌∷𝐵𝐸∶𝐵𝐷VI. 19 or ∷𝐴𝐵𝐸∶𝐴𝐵𝐷; therefore 𝑋𝐵𝑌=𝐴𝐵𝐷 932 If the interior and exterior vertical angels at 𝑃 of the triangle 𝐴𝑃𝐵 be bisected by straight lines which cut the base in 𝐶 and 𝐷, then the circle circumscribing 𝐶𝑃𝐷 gives the locus of the vertices of all triangles on the base 𝐴𝐵 whose sides 𝐴𝑃, 𝐵𝑃 are in a constant ratio.
Proof: The ∠𝐶𝑃𝐷=12(𝐴𝑃𝐵+𝐵𝑃𝐸) =a right angle therefore 𝑃 lies on the circumference of the circle, diameter 𝐶𝐷 (III 31). Also 𝐴𝑃∶𝐵𝑃∷𝐴𝐶∶𝐶𝐵∷𝐴𝐷∶𝐷𝐵 (VI 3 and A.) a fixed ratio. 933 𝐴𝐷 is divided harmonically in 𝐵 and 𝐶; i.e. 𝐴𝐷∶𝐷𝐵∷𝐴𝐶∶𝐶𝐵; or the whole line is to one extreme part as the other extreme part is to the middle part. If we put 𝑎, 𝑏, 𝑐 of the lengths 𝐴𝐷, 𝐵𝐷, 𝐶𝐷, the proportion is expressed algebraically by 𝑎∶𝑏∷𝑎−𝑐∶𝑐−𝑏, which is equivalent to 1𝑎+1𝑏=2𝑐 934 Also 𝐴𝑃∶𝐵𝑃=𝑂𝐴∶𝑂𝐶=𝑂𝐶∶𝑂𝐵 and 𝐴𝑃²∶𝐵𝑃²=𝑂𝐴∶𝑂𝐵VI.19 𝐴𝑃²−𝐴𝐶²∶𝐶𝑃²∶𝐵𝑃²−𝐵𝐶²VI.3 & 𝐵 935 If 𝑄 be the centre of the inscribed circle of the triangle 𝐴𝐵𝐶, and if 𝐴𝑄 produced meet the circumscribed circle, radius 𝑅, in 𝐹; and if 𝐹𝑂𝐺 be a diameter, and 𝐴𝐷 perpendicular to 𝐵𝐶; then 𝐹𝐶=𝐹𝑄=𝐹𝐵=2𝑅sin𝐴2i ∠𝐹𝐴𝐷=𝐹𝐴𝑂=𝐴2(𝐵−𝐶) and ∠𝐶𝐴𝐺=𝐴2(𝐵+𝐶)ii Proof of [i]: ∠𝐹𝑄𝐶=𝑄𝐶𝐴+𝑄𝐴𝐶 But 𝑄𝐴𝐶=𝑄𝐴𝐵=𝐵𝐶𝐹III.21 ∴𝐹𝑄𝐶=𝐹𝐶𝑄; ∴𝐹𝐶=𝐹𝑄; Similarly 𝐹𝐵=𝐹𝑄 Also ∠𝐺𝐶𝐹 is a right angle, and 𝐹𝐺𝐶=𝐹𝐴𝐶=12𝐴III.21 ∴𝐹𝐶=2𝑅sin𝐴2 936 If 𝑅, 𝑟 be the radii of the circumscribed and inscribed circles of the triangle 𝐴𝐵𝐶 (see last figure), and 𝑂, 𝑄 the centres; then 𝑂𝑄²=𝑅²−2𝑅𝑟 Proof: Draw 𝑄𝐻 perpendicular to 𝐴𝐶; then 𝑄𝐻=𝑟. By the isosceles triangle 𝐴𝑂𝐹, 𝑂𝑄²=𝑅²−𝐴𝑄⋅𝑄𝐹 (922, iii), and 𝑄𝐹=𝐹𝐶 (935,i), and by similar triangles 𝐺𝐹𝐶, 𝐴𝑄𝐻, 𝐴𝑄∶𝑄𝐻∷𝐺𝐹∶𝐹𝐶; therefore 𝐴𝑄⋅𝐹𝐶=𝐺𝐹⋅𝑄𝐻=2𝑅𝑟

Sources and References

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