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| − | {{WotWAnnounce|week=June 6-12}}
| + | #REDIRECT[[Angle bisector theorem]] |
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| − | == Introduction ==
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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.
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| − | <center>[[Image:Anglebisector.png]]</center>
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| − | == Proof ==
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| − | ===Method 1 ===
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| − | Because of the [[ratio]]s and equal [[angle]]s in the theorem, we think of [[similarity | 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:
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| − | <center>[[image:Anglebisectortheorem.PNG]]</center>
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| − | Since AB and CE are parallel, we know that <math> \angle BAE=\angle CEA </math> and <math> \angle BCE=\angle ABC </math>. Triangle ACE is [[isosceles triangle | isosceles]], meaning that AC = CE.
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| − | By [[AA similarity]], <math> \triangle DAB \cong \triangle DEC </math>. By the properties of similar triangles, we arrive at our desired result:
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| − | <center><math> \frac cm = \frac bn.</math> </center>
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| − | === Method 2 ===
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| − | Let <math> AD = d </math>. Now, we can express the area of triangle ABD in two ways:
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| − | <center><math> [ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB. </math></center>
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| − | Thus, <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm </math>.
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| − | Likewise, triangle ACD can be expressed in two different ways:
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| − | <center><math> [ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC. </math></center>
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| − | Thus, <math> \frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn</math>.
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| − | But <math> \angle CAD \cong \angle BAD </math> and <math> \sin \angle ADC = \sin \angle ADB </math> since <math> \angle ADC = \pi - \angle ADB </math>. Therefore, we can substitute back into our previous equation to get <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac bn </math>.
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| − | We conclude that <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn </math>, which was what we wanted.
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| − | In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
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| − | == Examples ==
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| − | #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>. | |
| − | # In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:''''' First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>. Thus, AP is the angle bisector of angle A, making our answer 0.
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| − | # Part '''(b)''', [[1959 IMO Problems/Problem 5]].
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| − | == See also ==
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| − | * [[Angle bisector]]
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| − | * [[Geometry]]
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| − | * [[Stewart's Theorem]]
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| − | [[Category:Geometry]]
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| − | [[Category:Theorems]]
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