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 Math Help > Geometry > Polygons and Triangles > Urquhart's Theorem

### Urquhartâ€™s Theorem:

Let OA and OB be two lines which intersect at O.
Let A' be a point on OA and B' be a point on OB.
Denote the intersection of AB' and A'B as O'.
Then OA + AO' = OB + BO' if and only if OA' + A'O' = OB' + B'O'.

Proof:

In what follows, we will ignore the special cases to be considered where we would otherwise be dividing by zero.

Define angles α = AOO', β = OO'A, θ = O'OB, φ = OO'B.

Thus OAO' = 180−α−β, OBO' = 180−θ −φ and by the law of sines:

OA = OO' sinβ / sin(180 − (α + β)) = OO' sinβ / sin(α + β)
O'A = OO' sinα / sin(180 − (α + β)) = OO' sinα / sin(α + β)
OB = OO' sinφ / sin(180 − (θ + φ)) = OO' sinφ / sin(θ + φ)
O'B = OO' sinθ / sin(180 − (θ + φ)) = OO' sinθ / sin(θ + φ)

Consequently

OA + AO' = OB + BO'
sinβ / sin(α + β) + sinα / sin(α + β) = sinφ / sin(θ + φ) + sinθ / sin(θ + φ)
(sinβ + sinα) / sin(α + β) = (sinφ + sinθ) / sin(θ + φ)
2sin((α+β)/2)cos((α−β)/2) / (2sin((α+β)/2)cos((α+β)/2)) = 2sin((θ+φ)/2)cos((θ−φ)/2) / (2sin((θ+φ)/2)cos((θ+φ)/2)
cos((α−β)/2)/cos((α+β)/2) = cos((θ−φ)/2)/cos((θ+φ)/2)

However, for arbitrary a, b, t, u we have the identity (provable by expanding both sides)

cos(a−b) cos(t+ u) − cos(a+b) cos(t−u) = sin(u+a) sin(b−t) − sin(u−a) sin(b+ t)

Thus the LHS is zero iff the RHS is zero, in other words

cos(a−b) cos(t+ u) = cos(a+b) cos(t−u)

is equivalent to

sin(u+a) sin(b−t) = sin(u−a) sin(b+ t)

Substituting a = α/2, b = β/2, t = θ/2, u = φ/2 and dividing through gives

cos((α−β)/2)/cos((α+β)/2) = cos((θ−φ)/2) cos((θ+φ)/2)
sin((φ+α)/2)/sin((φ−α)/2) = sin((β+θ)/2) sin((β−θ)/2)

Manipulating the last equation implies:

2sin((φ+α)/2) cos((φ−α)/2) / (2sin((φ−α)/2) cos((φ−α)/2)) = 2sin((β+θ)/2) cos((β−θ)/2) / (2sin((β−θ)/2) cos((β−θ)/2))
(sinφ + sinα) / sin(φ − α) = (sinβ + sinθ) / sin(β − θ)
sinφ / sin(φ − α) + sinα / sin(φ − α) = sinβ / sin(β − θ) + sinθ / sin(β − θ)

Finally, in considering triangles OA'O' and OB'O' we have OA'O' = φ−α, OB'O' = β − θ, OO'A' = 180 − φ, OO'B' = 180 − β. Therefore, by the law
of sines:

OA' = OO' sin(180 − φ) / sin(φ − α) = OO' sinφ / sin(φ − α)
O'A' = OO' sinα / sin(φ − α)
OB' = OO' sin(180 − β) / sin(β − θ) = OO' sin(β) / sin(β − θ)
O'B' = OO' sinθ / sin(β − θ)

and hence

sinφ / sin(φ − α) + sinα / sin(φ − α) = sinβ / sin(β − θ) + sinθ / sin(β − θ)
OA' + A'O' = OB' + B'O'

### Internet references

Mathworld: Urquhart's Theorem

Cut-the-knot: Urquhart's Theorem and Urquhart Elementary Proof

### Related pages in this website

Trig Equivalences: A particular Cosine Product Difference equals a Sin Product Difference, a wild and crazy trig identity, which is used here to prove Urquhart's Theorem.

The webmaster and author of this Math Help site is Graeme McRae.