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| − | {{WotWAnnounce|week=June 6-12}}
| + | #REDIRECT[[Angle bisector theorem]] |
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| − | == Introduction & Formulas ==
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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>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.
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| − | Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math>
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| − | <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>
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| − | == Proof ==
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| − | By <math>LoS</math> on <math>\angle ACD</math> and <math>\angle ABD</math>,
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| − | <math>\frac{AB}{BD}=\frac{sin(BDA)}{sin(BAD)}</math> ... <math>(1)</math> and
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| − | <math>\frac{AC}{AD}=\frac{sin(ADC)}{sin(DAC)}</math>... <math>(2)</math>
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| − | Well, we also know that <math>\angle BDA</math> and <math>\angle ADC</math> add to <math>180^\circ</math>. I think that means that we can use <math>sin(180-x)=sin(x)</math> here. Doing so, we see that <math>sin(BDA)=sin(ADC)</math>
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| − | I noticed that these are the numerators of <math>(1)</math> and <math>(2)</math> respectively. Since <math>\angle BAD</math> and <math>\angle DAC</math> are equal, then you get the equation for the bisector angle theorem.
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| − | == Examples & Problems ==
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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>.
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| − | # 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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