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Concylic Feet Part 2 (Posted on 2012-05-24) |
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DEFINITION
If P, Q, R, and S are distinct points, then
let (PQ&RS) denote the line through the
midpoint of line segment PQ and
perpendicular to line RS.
PROVE THE FOLLOWING
If A, B, C, and D are distinct points on a
circle, then (AB&CD), (AC&BD), and
(AD&CB) are concurrent.
Solution
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Comment 2 of 2 |
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Let a, b, c, d, be the position vectors of A, B, C, D, relative to the centre, O, of the circle, and let P be a point with position vector p = 0.5(a + b + c + d).
Let M be the mid-point of AB, so that the vector MP can be written as:
MP = p OM = 0.5(a + b + c + d) 0.5(a + b) = 0.5(c + d)
Then MP.CD = 0.5(c + d).(d c) = 0.5(d2 c2) = 0 since |c| = |d|.
So MP and CD are perpendicular and MP must be the line (AB&CD).
Since p is symmetrical in a, b, c and d, it follows that P will lie on all of the following lines, and therefore that they are all concurrent:
(AB&CD), (AC&BD), (AD&BC), (BC&AD), (BD&AC), (CD&AB).
It also confirms Jers discovery, since p 0.5(a + b + c) = 0.5d, so that as D moves on the circle, with A, B and C fixed, P describes a circle with half that radius, and centre at 0.5(a + b + c). This will be concentric with the original circle only when a + b + c = 0; i.e. when triangle ABC is equilateral.
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Posted by Harry
on 2012-05-25 23:07:51 |
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