We will be using the following theorem by Stewart (ST)
in our proof:
If P is a point on side MN of ΔLMN, then
|MN|•|LP|2 = |MP|•|LN|2 + |PN|•|LM|2 - |MN|•|MP|•|PN|
Let Q and R be the perpendicular projections of P onto sides AC and AB
respectively. Clearly, ARPQ is a cyclic quadrilateral with circumcircle Γ.
Let (X,Y) be an arbitrary pair from PAIRS with X and Y inside Γ and midpoint M.
Applying (ST) to cevian QY in ΔQRA,
|AR|•|QY|2 = |AY|•|QR|2 + |RY|•|AQ|2 - |AR|•|AY|•|RY|
= (|AR| - |RY|)•|QR|2 + |RY|•|AQ|2 - |AR|•(|AR| - |RY|)•|RY|
= |AR|•|QR|2 - |RY|•|QR|2 + |RY|•|AQ|2 - |RY|•|AR|2 + |AR|•|RY|2
= |RY|•[|AQ|2 - |AR|2 - |QR|2] + |AR|•[|QR|2 + |RY|2]
= |RY|•[-2•|AR|•|QR|•cos(∠ARQ)] + |AR|•[|QR|2 + |RY|2]
= |AR|•[-2•|QR|•|RY|•cos(∠ARQ) + |QR|2 + |RY|2]
= |AR|•[-2•|QR|•|RY|•cos(∠APQ) + |QR|2 + |RY|2]
= |AR|•[-2•|QR|•|RY|•(|PQ|/|AP|) + |QR|2 + |RY|2]
|QY|2 = -2•|QR|•|RY|•(|PQ|/|AP|) + |QR|2 + |RY|2 (1)
Swapping letters Q with R and X with Y in the above argument gives,
|RX|2 = -2•|QR|•|QX|•(|PR|/|AP|) + |QR|2 + |QX|2 (2)
Since ∠PXQ = ∠PXC = ∠PYB = ∠PYR,
|PQ|/|QX| = tan(∠PXQ) = tan(∠PYR) = |PR|/|RY| (3)
Combining (1), (2), and (3) gives,
|QX|2 + |QY|2 = |RX|2 + |RY|2 (4)
If points X and Y are outside Γ a similar argument gives
the same result as (4).
Applying (ST) to cevian QM in ΔQXY,
|XY|•|QM|2 = |MX|•|QY|2 + |MY|•|QX|2 - |XY|•|MX|•|MY|
or
|QM|2 = (|QY|2 + |QX|2)/2 - |MX|•|MY| (5)
Applying (ST) to cevian RM in ΔRXY,
|XY|•|RM|2 = |MX|•|RY|2 + |MY|•|RX|2 - |XY|•|MX|•|MY|
or
|RM|2 = (|RY|2 + |RX|2)/2 - |MX|•|MY| (6)
Combining (4), (5), and (6) gives,
|QM| = |RM|
Therefore, the midpoint M lies on the perpendicular bisector of QR.
QED
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