The following particular cases of (iv) are of interest. For the ortho-centre, in-centre, centroid, and symmedian point respectively

Sin A cos λ + sin B cos μ + sin C cos ν = o

cos A/2 cos λ + cos B/2 cos μ + cos C/2 cos ν = o

m₁ cos λ + m₂ cos μ + m₃ cos ν = o

am₁ cos λ + bm₂ cos μ + cm₃ cos ν = o,

where m₁, m₂, and m₃ are the medians of the triangle ABC.

If O be either of the two points whose pedal triangles, are equiangular, then, since for these points a_ρ1 = b_ρ2 = c_ρ3,

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a √X_oα_oα + b √Y_oβ_oβ + c √Z_o γ_oγ = o.

If O be the focus of a conic inscribed in the triangle ABC, then the equation of the conic is

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a √X_oα_oα + b √Y_oβ_oβ + c √Z_o γ_oγ = o.

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Comparing this with the equation of the maximum inscribed ellipse, √ aα + √bβ √γ = o, we have aX_oα_o = bY_oβ_o = cZ_oγ_o, whence

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cosλ/AO + cosμ/BO + cosν/CO = o.

For the Brocard ellipse this gives

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cosλ/BC. AO + cosμ/CA. BO + cosν/AB. CO = o.

Let U¹ be the result of substituting in U the co-ordinates (X¹, Y¹, Z¹) for (X, Y, Z). Suppose a circle of radius p¹ concentric with U to pass through the point (X¹Y¹Z¹), then

U¹ = aα_oX¹ + bγ_oY¹ − 2RS_o − 2δ ρ²

O = aα_oX¹ + bβ_oY¹ − 2RS_o − 2δ ρ^{1 2}

∴ U¹ = 2δ (ρ^{1 2} − ρ²) = 2δ t^{1 2},

where t¹ is the length of the tangent from (X¹Y¹Z¹) to the circle U = O.

Let t₁, t₂, and t₃ be the lengths of the tangents to the circle U from A, B, C respectively: then

2 δ t²₁ = bβ_oc² + cγ_ob² − 2 δ (R² − d² + ρ²);

i.e., 2 δ (t₁² + R² − d₂ + ρ²) = bβ_oc² + cγ_ob².

We have X_o = t₁² + ρ², whence, if OH subtend at A, B, and C the angles θ, φ ψ respectively,

abc ρ¹ cos θ = bβ_oc² + cγ_ob²

abc ρ² cos φ = aα_oc² + cγ_oa²

abc ρ¹ cos ψ = α_ob² + bβ_oa²

2 δ = aα_o + bβ_o + cγ_o.

Eliminating aα_o bβ_o cγ_o we have

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which reduces to the relation

aρ_o cos A θ + bρ₂ cos B cos φ + cρ₃ cos C cosψ = 2 δ.

From the equations

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αρ₁ cos θ/bc = bβ_o/b + γ_o/c, bρ₂ cosφca = γ_o/o + α_o/a, cρ₃ cos ψ/ab = α_o/a + β_o/b,

we see that if O lie on the trilinear polar of the symmedian point BC² AO cosθ + CA²BO cosφ + AB²CO cosψ = o.

Solving for α_o, β_o, γ_o from the above equations, and substituting in the equation of the circle ABC, we obtain the relation

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BC √OA cosθ + CA √OBcosφ + AB√OC cosψ = o

for any point O on the circle.

If t_o be the length of the tangent to any circle from the middle point of BC, then

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2 δ t²₂ = aα_oc² + cγ_oa²−2 (RS_o + δρ²)

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2 δ t²₃ = aα_ob² + bβ_oa²−2 (RS_o + δρ²)

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2 δ t²_o = aα_om²₁ + a²/4(bβ_o + cγ_o)−2(RS_o + δρ²)

where m₁ is the median drawn through A.

Hence 2δ(t²₂ + t²₃−2t²_o

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= aα_o(b² + c²−2m²₁) + a²/2(bβ_o + cγ_o)

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= a²/2 δ;

i.e. t²₂ + t²₃ = 2t²_o + a²/2,

an extension of the Theorem of Apollonius.

If U be the polar circle of the triangle ABC, then

U ≡ tan AX + tan BY + tan CZ − 2 δ = o

∴ δ t²₂ = κ (c² tan A + a² tan C − 2δ),

where κ = 4R²cos A cos B cos C.

which reduces to t₂² = ca cos B. Also t₃² = ab cos C, hence t₃₂ + t₃² = a².

If in the triangle ABC the angle A be obtuse, then the sum of the squares of the tangents to the polar circle from B and C is BC².

If U = o pass through the point (X₁Y₁Z₁), then the locus of O (α_oβ_oγ_o) is the circle

V ≡ aαX₁ + bβY₁ + cγZ₁ − 2RS −2δρ² = o.

If t₁, t₂, t₃ be the lengths of the tangents to the circle V from A, B, C respectively, then X₁ = t₁² + p², Y₁ = t₂² + p², Z = t₃² + p², and the equation of the circle takes the form

V ≡ (t²₁aα + t²₂bβ + cγ) (aα + bβ + cγ) − abc S = o.

If V touch the circle ABC, then expressing that the radical axis of the two circles is a tangent to S we have

at₁ ± bt₂± ct₃ = o.

from which we obtain Ptolemy's theorem if we suppose V to reduce to a point-circle

This extension of Ptolemy's theorem may be proved geometrically as follows:—

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Suppose the circle V to touch the circle ABC at the point O on the arc BC, and let AO, BO, CO meet V in the points D, E, F respectively; then t₂² = BO. BE and t₃² = CO. CF. Also OE/OF = OB/OC, and therefore BE/CF = OB/OC, hence t₂²: t₃² = OB²: OC²,

i.e., t₁: t₂: t₃ = OA: OB: OC.

By Ptolemy's theorem BC . OA = AC . OB + AB . OC;

i.e., at₁ = bt₂ + ct₃.

If t be the length of the tangent to the circle V = o from any point P (α_oβ_oγ_o), then

2 δ t² = aα_ot₁² + bβ_ot₂² + cγ_ot₈² − 2RS_o.

For the circle BPC, t = t₂ = t₃ = o.

Consider now the circles BPC, CPA, APB. We have

aα_ot²₁ = bβ_ot²₂ = cγ_ot²₃ = 2RS_o

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Hence aα_o + bβ_o − cγ_o/2RS_o = 1/t²₁ + 1/t²₂ + 1/t₃²

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i.e. 1/t²₁ + 1/t²₂ + 1/t²₃ = 1/R²−d²

where d is the distance of P from the circum-centre of the triangle ABC.

If P be the symmedian point of the triangle ABC, then at₁ = bt₂ = ct₃: if G be the centroid of the triangle t₁² = t₂² = t₃² = ⅓ (a²)

If the circle V reduce to a point P₁ whose tripolar co-ordinates are (X₁Y₁Z₁) we have

X₁aα + Y₁bβ + Z₁cγ_o = 2RS.

For a point-circle at P₂ (X₂Y₂Z₂)

X₂aα + Y₂bβ + Z₂cγ = 2RS.

Hence the radical axis of the pair of circles is

(X₁ − X₂) aα + (Y₁ −Y₂) bβ + (Z₁ − Z₂) cγ = o.

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If P _l, P₂ be inverse points determined by the equations X/l = Y/m = Z/n, then X₁ − X₂/l = Y₁ − Y₂/m = Z₁ − Z₂/n = κ₁ − κ₂, where κ₁ and κ₂ are the roots of the equation κ² [Σ (a²l²) −2 Σ (bc cos Amn]

− 2κ abc [(a cos Al)] + a²b²c² = o.

Hence the equation of the line bisecting perpendicularly the above pair of inverse points is

laα + mbβ + ncγ = o.

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The two points whose pedal triangles are equilateral are determined by the equations a²X = b²Y = c²Z. Hence these points are equidistant from the line α/a + β/b + γ/c = o, which is therefore perpendicular to the Brocard diameter Σ [a² (b² – c²) X] = o, on which the points lie.

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The two points whose distances from A, B, C are proportional respectively to a, b, c are given by the equations X/a² = Y/b² = Z/c²: they therefore lie on Euler's line, Σ[(b² − c²) X] = o, and are equidistant from the line a³α + b³β + c³γ = o.

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Let two pairs of inverse points be determined by the equations X/l₁ = Y/m₁ = Z/n₁ and X/l₂ = Y/m₂ = Z/n₂; the centre of the circle through them will be determined by the equations

l₁aα + m₁bβ + n₁cγ = o

l₂aα + m₂bβ + n₂cγ = o,

whence aα : bβ : cγ = m₁n₂ −m₂n₁ : n₁l₂ = n₂l₁ : l₁m₂ − l₂m₁, and the equation of the circle will be

(m₁n₂ −m₂n₁) × + (n₁l₂ − n₂l₁) Y + (l₁m₂ − l₂m₁) Z = o.

The two pairs of points will lie respectively on the diameters

(m₁ − n₁) × + (n₁ − l₁) Y + (l₁ − m₁) Z = o

(m₂ − n₂) × + (n₂ − l₂) Y + (l₂ − m₂) Z = o

We now proceed to find the equation in tripolar co-ordinates of the inverse of the circle U ≡ o with respect to S, the circum-circle of the triangle ABC.

U ≡ lX + mY + nZ − h² = o

S ≡ a cos AX + b cos BY + c cos CZ − abc = o.

Let P be any point on U and Q its inverse. Let (X, Y, Z), (X¹, Y¹, Z¹) be the tripolar co-ordinates of P and Q respectively, then

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X/X¹ = Y/Y¹ = Z/Z¹ = HP²/R²,

where H is the circum-centre.

Hence HP² (lX¹ + mY¹ + nZ¹) = h²R².