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# Similar triangles Example

Question: In the figure above, ABC is a triangle. BD bisects the angle ABC. Prove AB/BC = AD/DC.
Proof:
Extended BD, and draw a line passing through point C and parallel to AB, the two lines intersects at point E. Since BA // CE (Made), so angle 1 = angle 3 (Two parallel lines BA and CE are cut by BE, then their alternate interior angles 1 and 3 are equal.)
Since angle 1 = angle 2 (Given), and angle 1 = angle 3, so angle 2 = angle 3.
In triangle BCE, since angle 2 = angle 3, so BC = CE (Property of an isosceles triangle.)
In triangle ADB and triangle CDE,
since angle 1 = angle 3, and angle ADB = angle CDE (Vertical angles are equal.)
so triangle ADB is similar to triangle CDE (AA Theorem)
so AB/CE = AD/DC (In similar triangles, corresponding sides are in proportion.)
Since CE = BC, so AB/BC = AD/DC

Note: In similar triangles ADB and CDE, AD opposite angle 1 and DC opposite angle 3, since angle 1 = angle 3, so AB and CE are a pair of corresponding sides. Look another pair of corresponding sides, AB opposite angle ADB and CE opposite angle CDE, since angle ADB = angle CDE, so AB and CE are a pair of corresponding sides. In two similar triangles, corresponding sides are in proportion. That is, AB/CE = AD/DC.

What does this example tell us? In a triangle, the two line segments cut by an angle bisector and another two sides are in proportion.

Use this example to help middle school students learning inference.