Elementary Geometry

Miscellaneous Propositions

920 To find the point in a given line 𝑄𝑌, the sum of whose distances from two fixed points 𝑆, 𝑆′ is a minimum.
Draw 𝑆𝑌𝑅 at right angles to 𝑄𝑌, making 𝑌𝑅=𝑌𝑆. Join 𝑅𝑆′, cutting 𝑄𝑌 in 𝑃. Then 𝑃 will be the required point.
Proof: For, if 𝐷 be any other point on the line, 𝑆𝐷=𝐷𝑅 and 𝑆𝑃=𝑃𝑅. But 𝑅𝐷+𝐷𝑆′ is >𝑅𝑆′; therefore, ⋯ 𝑅 is called the reflection of the point 𝑆, and 𝑆𝑃𝑆′ is the path of a ray of light reflected at the line 𝑄𝑌.
If 𝑆, 𝑆′ and 𝑄𝑌 are not in the same plane, make 𝑆𝑌, 𝑌𝑅 equal perpendiculars as before, but the last in the plane of 𝑆′ and 𝑄𝑌.
Similarly, the point 𝑄 in the given line, the difference of whose distances from the fixed points 𝑆 and 𝑅′ is a maximum, is found by a like construction.
The minimum sum of distances from 𝑆, 𝑆′ is given by (𝑆𝑃+𝑆′𝑃)²=𝑆𝑆′²+4𝑆𝑌⋅𝑆′𝑌′ And the maximum difference from 𝑆 and 𝑅′ is given by (𝑆𝑄+𝑅′𝑄)²=(𝑆𝑅′)²−4𝑆𝑌⋅𝑅′𝑌′ Proved by VI. D., since 𝑆𝑅𝑅′𝑆′ can be inscribed in a circle. 921 Hence, to find the shortest distance from 𝑃 to 𝑄 en route of the lines 𝐴𝐵, 𝐵𝐶, 𝐶𝐷; in other words, the path of the ray reflected at the successive surfaces 𝐴𝐵, 𝐵𝐶, 𝐶𝐷.
𝑎𝑃₂ cutting 𝐵𝐶 in 𝑏. Join 𝑏𝑃₁ cuting 𝐴𝐵 in 𝑐. Join 𝑐𝑃. 𝑃 Find 𝑃₁, the reflection of 𝑃 at the first surface; then 𝑃₂, the reflection of 𝑃₁ at the second surface; next 𝑃₃, the reflection of 𝑃₂ at the third surface; and so on if there be more surfaces. Lastly, join 𝑄 with 𝑃₃, the last reflection, cutting 𝐶𝐷 in 𝑎. Join 𝑎𝑃₂ cutting 𝐵𝐶 in 𝑏. Join 𝑏𝑃₁ cuting 𝐴𝐵 in 𝑐. Join 𝑐𝑃. 𝑃𝑐𝑏𝑎𝑄 is the path requied.
The same construction will give the path when the surfaces are not, as in the case considered, all perpendicular to the same plane. 922 If the straight line 𝑑 from the vertex of a triangle divide the base into segments 𝑝, 𝑞, and if ℎ be the distance from the point of section to the foot of the perendicular from the vertex on the base, then 𝑏²+𝑐²=𝑝²+𝑞²+2𝑑²+2ℎ(𝑝−𝑞)II. 12, 13 The following cass are important:

1. When 𝑝=𝑞, 𝑏²+𝑐²=2𝑞²+2𝑑² i.e., the sum of the squares of two sides of a triangle is equal to twice the square of half the base, together with twice the square of the bisectin line drawn from the vertex.
2. When 𝑝=2𝑞, 2𝑏²+𝑐²=6𝑞²+3𝑑²II. 12, 13
3. When the triangle is isosceles, 𝑏²=𝑐²=𝑝𝑞+𝑑²

923 If 𝑂 be the centre of an equilateral triangle 𝐴𝐵𝐶 and 𝑃 any point in space. Then 𝑃𝐴²+𝑃𝐵²+𝑃𝐶²=3(𝑃𝑂²+𝑂𝐴²) Proof: 𝑃𝐵²+𝑃𝐶²=2𝑃𝐷²+2𝐵𝐷²922, i Also 𝑃𝐴²+2𝑃𝐷²=6𝑂𝐷²+3𝑃𝑂²922, ii and 𝐵𝑂=2𝑂𝐷 therefore ⋯

Cor:

Hence, if 𝑃 be any point on the surface of a sphere, centre 𝑂, the sum of the squares of its distances from 𝐴, 𝐵, 𝐶 is constant. And if 𝑟, the radius of the sphere, be equal to 𝑂𝐴, the sum of the same squares is equal to 6𝑟². 924 The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals plus four times the square of the line joining the middle points of the diagonals.  922, i 925

Cor

The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals.

Sources and References

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