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| Split an Altitude (Posted on 2010-09-24) |
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Let BB' be an altitude of right triangle ABC
(where B is the right angle).
Prove that the common chord of the circle with
center B and radius |BB'| and the circumcircle
of triangle ABC bisects BB'.
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Submitted by Bractals
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Rating: 4.0000 (1 votes)
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Solution:
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(Hide)
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Let O be the midpoint of side AC (also the
center of the circumcircle of triangle ABC).
Let the line BO intersect the circumcircle
again at point R. Let the common chord PQ
intersect lines BB' and BO at points M and
N respectively. Construct line segments BP
and PR.
From similar right triangles BPR and BNP,
|BP|2 = |BR||BN|
or
|BB'|2 = 2|BO||BN| (1)
From similar right triangles BNM and BB'O,
|BO||BN| = |BB'||BM| (2)
Combining (1) and (2) gives
|BB'| = 2|BM|.
Therefore, PQ bisects BB'.
Note: See Harry's post for an alternate solution.
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