Difference between revisions of "Angle Bisector Theorem"
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The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. Likewise, the converse of this theorem holds as well. | The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. Likewise, the converse of this theorem holds as well. | ||
− | <asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D, | + | <asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy> |
== Proof == | == Proof == |
Revision as of 15:06, 26 June 2016
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Introduction
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . Likewise, the converse of this theorem holds as well.
Proof
Method 1
Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:
Since AB and CE are parallel, we know that and . Triangle ACE is isosceles, with AC = CE.
By AA similarity, . By the properties of similar triangles, we arrive at our desired result:
Method 2
Since B,D, and C are collinear, . Now consider the perpendicular from to and to . Every point on the angle bisector of an angle is equidistant to the sides of the angle, so the height to is equal to the height to . Thus . Thus , so . We can prove the converse by the Phantom Point Method, since we can find and in terms of , , and , and prove that the points are the same.
Method 3
Let . Now, we can express the area of triangle ABD in two ways:
Thus, .
Likewise, triangle ACD can be expressed in two different ways:
Thus, .
But and since . Therefore, we can substitute back into our previous equation to get .
We conclude that , which was what we wanted.
In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
Examples
- Let ABC be a triangle with angle bisector AD with D on line segment BC. If and , find AB and AC.
Solution: By the angle bisector theorem, or . Plugging this into and solving for AC gives . We can plug this back in to find . - In triangle ABC, let P be a point on BC and let . Find the value of .
Solution: First, we notice that . Thus, AP is the angle bisector of angle A, making our answer 0. - Part (b), 1959 IMO Problems/Problem 5.