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Show the internal bisector of one angle of a triangle and the external bisectors of the other two external angles meet at a single point and the perpendicular distance to the the side opposite the internal angle and to the other two sides extended is equal.
Show the internal bisector of one angle of a triangle and the external bisectors of the other two external angles meet at a single point and the perpendicular distance to the the side opposite the internal angle and to the other two sides extended is equal.
Draw ∇ABC
Extend BA to D and BC to E
Construct bisectors of ^GAC and ^ECA which intersect at F
Construct perpendicular from BD to F, meeting BD at G
Construct perpendicular from BE to F, meeting BE at H
Construct perpendicular from AC to F, meeting AC at J
Consider ∇AFG and ∇AFJ
^GAF = ^JAF (Construct)
^AGF = 90º = ^AJF (Construct)
∴ ^AFG = ^AFJ (Angles of triangle equal 2 rt. angles.)
AF = AF (common)
∴ ∇AGF = ∇AJF (2 angles and inc. side equal)
∴ GF = JF
Similarly considering ∇CJF and ∇CHF
JF = HF
∴ GF = JF = HF
Join B to F
Consider ∇BGF and ∇BHF
BF = BF (Common)
^BGF = 90º = ^BHF (Construct)
GF = HF (Proven above)
∴ ∇BGF = ∇BHF (Rt. ang. triangle, hypotenuse and 1 side equal)
∴ ^GBF = ^HBF
Therefore all three bisectors meet at a single point equidistant from the three sides.
Therefore all three bisectors meet at a single point equidistant from the three sides.
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