Symmetrical Point Projection Formula

Theorem: Symmetrical Point Projection Formula

       Let Q₁ and Q₂ be points on the same longitude of the sphere and
     symmetrical about the equator. Let the stereographic projections of
     these points be P₁and P₂ respectively. If S is the South Pole of the
     sphere and T is the image of the point on the equator halfway between
     Q₁ and Q₂, then ST²=SP₁* SP₂.   (Refer to Figure 1)

Figure 1

The plan is to show that triangles P₂SN and NSP₁ are similar.

This will allow us to do some simple algebra and prove that ST²=SP₁* SP₂.

Figure 2 This section will show that triangle NSP₁ and triangle NQ₁S are similar.
Referring to Figure 2:
    Angle SNQ₁ is the same angle as P₁NS.
    Angle NQ₁S is a right angle since it is inscribed in a semicircle.
    Angle NSP₁ is a right angle since SP₁ is tangent to the circle at point S and NS is the diameter of
    the circle.
    Therefore the measures of angle NQ₁S and angle NSP₁are the same.
    Since triangle NSP₁ and triangle NQ₁S are both right triangles, and each share another angle
    beside the right angle, the third angle in each triangle must have the same measures.
    Triangle NSP₁ and triangle NQ₁S are similar by AAA.

Figure 3 This section will show that triangle SQ₂N and triangle NQ₁S are similar.
Referring to Figure 3:
    Angle NQ₁S and angle SQ₂N are right angles since they are inscribed in a semicircle.
    Angle Q₁NS and angle Q₂SN are congruent since points Q₁ and Q₂ are at the same
    longitude on the sphere and symmetrical about the equator. (i.e. the arcs they are inscribed in
    have the same measure)
    Since triangle SQ₂N and triangle NQ₁S are both right triangles, and each share another angle
    beside the right angle, the third angle in each triangle must have the same measures.
    Triangle SQ₂N and triangle NQ₁S are similar by AAA.

Figure 4 This section will show that triangle SQ₂N and triangle P₂SN are similar.
Referring to Figure 4:
    Angle SNQ₂ is the same angle as P₂NS.
    Angle SQ₂N is a right angle since it is inscribed in a semicircle.
    Angle P₂SN is a right angle since SP₂ is tangent to the circle at point S and NS is the diameter of
    the circle.
    Therefore the measures of angle SQ₂N and angle P₂SN are the same.
    Since triangle SQ₂N and triangle P₂SN are both right triangles, and each share another angle
    beside the right angle, the third angle in each triangle must have the same measures.
    Triangle SQ₂N and triangle P₂SN are similar by AAA.

Since triangle NSP₁ is similar to triangle NQ₁S, and triangle NQ₁S is similar to triangle SQ₂N, and triangle SQ₂N is similar to triangle P₂SN, triangle P₂SN and NSP₁ must be similar.

Figure 5 Referring to Figure 5: Using the ratios of the corresponding sides: SP₂/NS = NS/SP₁ (NS)² = (SP₁)(SP₂) From the Equator Projection Theorem we know that 2CE = ST. CE= NS/2 NS = ST Substituting ST for NS gives us: (ST)² = (SP₁)(SP₂) Q.E.D.