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| Midpoint of Two Chords (Posted on 2013-09-15) |
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Let AB and BC be chords of a circle ( with
|AB| > |BC| ). Let M be the midpoint of
arc ABC and F the foot of the perpendicular
from M to AB.
Prove that |AF| = ( |AB| + |BC| )/2.
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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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Construct point D on chord AB such that |AD| = |BC|.
Construct chords MA, MB, and MC. Construct line
segment MD.
Angles MAB and MCB are equal since they are subtended
by the same chord MB. Chords MA and MC are equal
since the arcs MA and MC are equal. Therefore,
triangles MAD and MCB are congruent by SAS. Therefore,
|MD| = |MB|. Right triangles MFD and MFB have a common
side MF and equal hypotenuses, therefore the third
sides DF and FB are equal.
|AB| + |BC| = |AD| + |DF| + |FB| + |BC|
= |AD| + |DF| + |DF| + |AD|
= 2(|AD| + |DF|)
= 2|AF|
QED
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Subject |
Author |
Date |
 | Solution | Harry | 2013-09-20 16:51:33 |
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