Problem: Given any triangle ABC, construct an angle bisector from the given angle, thru the opposite side of the triangle. Prove that a/b=x/y. See GSP link for illustration.

First, we want to begin by constructing a parallel to segment CD thru B and extending AC to point E where BE and AE intersect. Click here to view.

Now we need to show that triangle ACD is similar to triangle AEB.

Angle A is congruent to itself.

Angle EBA is congruent to angle CDA because CD is parallel to EB (by construction), both angles are cut by the same transversal.

By similar reasoning, angle BEA is congruent to angle DCA.

From this, we can deduce that EC/CA=x/y, where b=CA. We now need to show that EC=a and the proof will be complete.

Using the construction above, we need to construct a parallel to BA thru C.

Now we can notice that FCDB is a parallelogram. Thus, angle FBC is congruent to angle BCD. Since CD was our original bisector, angle BCD is congruent to angle DCA.

From earlier, angle DCA was congruent to angle E, therefore angle EBC is congruent to angle E.

Now we have an isosceles triangle BCE, giving us a congruent to EC,

therefore EC/CA=x/y can as well be written as a/b=x/y.

For an illustration of this proof, click bisectors to see numerically how this proof works for any shape triangle.

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