Elementary Geometry

Miscellaneous Propositions

To Construct a Triangle from Certain Data

960 When amongst the data we have the sum or difference of the two sides 𝐴𝐵, 𝐴𝐶; or the sum of the segments of the base made by 𝐴𝐺, the bisector of the exterior vertical angle; or the difference of the segments made by 𝐴𝐹, the bisector of the interior vertical angle; the following construction will lead to the solution.
Make 𝐴𝐸=𝐴𝐷=𝐴𝐶. Draw 𝐷𝐻 parallel to 𝐴𝐹; and suppose 𝐸𝐾 drawn parallel to 𝐴𝐺 to meet the base produced in 𝐾; and complete the figure. Then 𝐵𝐸 is the sum, and 𝐵𝐷 is the difference of the sides.
𝐵𝐾 is the sum of the exterior segments of the base, and 𝐵𝐻 is the difference of the interior segments. ∠𝐵𝐷𝐻=𝐵𝐸𝐶=12𝐴, ∠𝐴𝐷𝐶=𝐸𝐴𝐺=12(𝐵+𝐶), ∠𝐷𝐶𝐵=12𝐷𝐹𝐵=12(𝐶−𝐵). 961 When the base and the vertical angle are given; the locus of the vertex is the circle 𝐴𝐵𝐶 in figure (935); and the locus of the centre of the inscribed circle is the circle, centre 𝐹 and radius 𝐹𝐵. When the ratio of the sides is given, see (932). 962 To construct a triangle when its form and the distances of its vertices from a point 𝐴′ are given.

Analysis

Let 𝐴𝐵𝐶 be the required triangle. On 𝐴′𝐵 make the triangle 𝐴′𝐵𝐶′ similar to 𝐴𝐵𝐶, so that 𝐴𝐵∶𝐴′𝐵∷𝐶𝐵∶𝐶′𝐵. The angles 𝐴𝐵𝐴′, 𝐶𝐵𝐶′ will also be equal ; therefore 𝐴𝐵∶𝐵𝐶∷𝐴𝐴′∶𝐶𝐶′, which gives 𝐶𝐶′, since the ratio 𝐴𝐵∶𝐵𝐶=𝐴′𝐵∶𝐵𝐶′ is known. Hence the point 𝐶 is found by constructing the triangle 𝐴′𝐶𝐶′. Thus 𝐵𝐶 is determined, and thence the triangle 𝐴𝐵𝐶 from the known angles. 963 To find the locus of a point 𝑃, the tangent from which to a given circle, centre 𝐴, has a constant ratio to its distance from a given point 𝐵.
Let 𝐴𝐾 be the radius of the circle, and 𝑝∶𝑞 the given ratio. On 𝐴𝐵 take 𝐴𝐶, a third proportional to 𝐴𝐵 and 𝐴𝐾, and make 𝐴𝐷∶𝐷𝐵=𝑝²∶𝑞² With centre 𝐷, and a radius equal to a mean proportional between 𝐷𝐵 and 𝐷𝐶, describe a circle. It will be the required locus.

Proof

Suppose 𝑃 to be a point on the required locus. Join 𝑃 with 𝐴, 𝐵, 𝐶, and 𝐷.
Describe a circle about 𝑃𝐵𝐶 cutting 𝐴𝑃 in 𝐹, and another about 𝐴𝐵𝑃 cutting 𝐵𝑃 in 𝐺, and join 𝐴𝐺 and 𝐵𝐹. Then 𝑃𝐾²=𝐴𝑃²−𝐴𝐾²=𝐴𝑃²−𝐵𝐴⋅𝐴𝐶(constr.)=𝐴𝑃²−𝑃𝐴⋅𝐴𝐹III.36 =𝐴𝑃⋅𝑃𝐹(II.2)=𝐺𝑃⋅𝑃𝐵(III.36) Therefore, by hypothesis, 𝑝²∶𝑞²=𝐺𝑃⋅𝑃𝐵∶𝑃𝐵²=𝐺𝑃⋅𝑃𝐵=𝐴𝐷∶𝐷𝐵(by constr.) therefore ∠𝐷𝑃𝐺=𝑃𝐺𝐴 (VI.2)=𝑃𝐹𝐵(III.22)=𝑃𝐶𝐵(III.21) Therefore the triangles 𝐷𝑃𝐵, 𝐷𝐶𝑃 are similar; therefore 𝐷𝑃 is a mean proportional to 𝐷𝐵 and 𝐷𝐶. Hence the construction. 964

Cor

if 𝑝=𝑞 the locus becomes the perpendicular bisector of 𝐵𝐶, as is otherwise shown in (1003). 965 To find the locus of a point 𝑃, the tangents from which to two given circles shall have a given ratio. (See also 1036)
Let 𝐴, 𝐵 be the centres, 𝑎, 𝑏 the radii (𝑎>𝑏), and 𝑝∶𝑞 the given ratio. Take 𝑐, so that 𝑐∶𝑏=𝑝∶𝑞, and describe a circle with centre 𝐴 and radius 𝐴𝑁=𝑎²−𝑐². Find the locus of 𝑃 by the last proposition, so that the tangent from 𝑃 to this circle may have the given ratio to 𝑃𝐵. It will be the required locus.

Proof

By hypothesis and construction, 𝑝²𝑞²=𝑃𝐾²𝑃𝑇²=𝑐²𝑏²=𝑃𝐾²+𝑐²𝑃𝑇²+𝑏²=𝐴𝑃²−𝑎²+𝑐²𝐵𝑃²=𝐴𝑃²−𝐴𝑁²𝐵𝑃²

Cor

Hence the point can be found on any curve from which the tangents to two circles shall have a given ratio. 966 To find the locus of the point from which tangents to two given circles are equal.
Since, in (965), we have now 𝑝=𝑞, therefore 𝑐=𝑏, the construction simlifies to the following:
Take 𝐴𝑁=𝑎²−𝑏², and in 𝐴𝐵 take 𝐴𝐵∶𝐴𝑁∶𝐴𝐶. The perpendicular bisector of 𝐵𝐶 is the required locus. But, if the circles intersect, then their common chord is at once the line required. See Radical Axis (985).

Sources and References

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