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This essential trains for: AMC-10, AMC-12, AIME.
The bisector of an interior angle of a triangle intersects the opposite side to form segments in a ratio equal to the ratio of the other two sides of the triangle.
Proof: like in any proof involving products or ratios of segments, we have to look for similar triangles.
Although there is a pair of congruent angles, there is no other element to help form similar triangles.
Use an additional construction to form similar triangles. Any construction will work. For example, make an angle congruent to ∠ABN at C:
Thanks to the additional construction, we can now find two congruent triangles: BAN and CAM.
Write the similarity ratios:
This looks very close to the result that we need, but is still different, since we have the length CM in it. The length CM is part of the additional construction.
The angles ∠ANB and ∠CMN are congruent since the angles of each triangle add up to 180°.
However, we notice that, since the angles ∠BNA and ∠MNC are vertical, they are congruent. By transitivity:
Therefore, the triangle CMN is isosceles and we have:
We can now substitute the length of the segment CN for the length of CM: