Interior angle bisector theorem
Submitted by Structure on Thu, 12/20/2007 - 20:33.
Let ABC be a triangle. Let AM be the bisector of the angle A. Then we have:
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Proof:
Let d be the line parallel at the line AM that passes through C. Let F be the intersection between this line d and AB.
AFC = BAM (because AM is parallel with FC)
BAC = MAC (because AM is the bisector of BAC)
MAC = ACF (because AM is parallel with FC)
Because AFC = ACF the triangle AFC is isoscele. So AF = AC.
From the Thales theorem we have:
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because AF = AC we have:
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