Home > Shapes > Geometry
| Incircle Bisector (Posted on 2010-10-23) |
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Let ABC be a triangle with /ABC < /ACB < 90°.
Let D be a point on side BC such that
|AD| = |AC|.
The incircle of triangle ABC is tangent to sides
AC and BC at points P and Q respectively. Let J
be the incenter of triangle ABD.
Prove that line PQ bisects line segment CJ.
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Submitted by Bractals
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Rating: 3.0000 (2 votes)
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Solution:
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Lemma without proof.
If the incircle of ΔXYZ kisses side XY at point Z',
then 2|XZ'| = |XY| + |XZ| - |YZ|.
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Let E be the foot of the perpendicular from J to BC
and F the intersection of BC and the line through J
parallel to PQ.
/JDE = 90° - ½/ADC =
90° - ½/ACQ = /PQC = /JFE.
Therefore, |FD| = 2|DE|. Using the lemma,
|FC| = |BC| - |BD| + |FD| = |BC| - |BD| + 2|DE|
= |BC| - |BD| + ( |DB| + |DA| - |AB| )
= |BC| + |DA| - |AB| = |CB| + |CA| - |AB|
= 2|CQ| = 2|QC|
Let M be the intersection of PQ and CJ. From
similar triangles FCJ and QCM,
|QC||CJ| = |FC||CM| = 2|QC||CM|
or
|CJ| = 2|CM|
Therefore, line PQ bisects line segment CJ.
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