Difference between revisions of "Angle Bisector Theorem"
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{{WotWAnnounce|week=June 6-12}} | {{WotWAnnounce|week=June 6-12}} | ||
− | == Introduction == | + | == Introduction & Formulas == |
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>. It follows that <math> \frac cb = \frac mn </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>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well. | ||
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== Proof == | == Proof == | ||
− | By | + | By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</math>, |
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− | == Examples == | + | <cmath>\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ |
+ | \frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}</cmath> | ||
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+ | First, because <math>\bar{AD}</math> is an angle bisector, we know that <math>m\angle BAD = m\angle CAD</math> and thus <math>\sin(BAD) = \sin(CAD)</math>, so the denominators are equal. | ||
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+ | Second, we observe that <math>m\angle BDA + m\angle CDA = \pi</math> and <math>\sin(\pi - \theta) = \sin(\theta)</math>. | ||
+ | Therefore, <math>\sin(BDA) = \sin(CDA)</math>, so the numerators are equal. | ||
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+ | It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}</cmath> | ||
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+ | == Examples & Problems == | ||
# Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>. | # Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>. |
Revision as of 11:27, 21 December 2020
This is an AoPSWiki Word of the Week for June 6-12 |
Introduction & Formulas
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . It follows that . Likewise, the converse of this theorem holds as well.
Further by combining with Stewart's Theorem it can be shown that
Proof
By the Law of Sines on and ,
First, because is an angle bisector, we know that and thus , so the denominators are equal.
Second, we observe that and . Therefore, , so the numerators are equal.
It then follows that
Examples & Problems
- 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.