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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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| − | <asy>
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| − | size(200);
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| − | defaultpen(fontsize(10));
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| − | real a,b,c,d;
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| − | pair A=(1,4), B=(-5,0), C=(3,0), D;
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| − | b = abs(C-A);
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| − | c = abs(B-A);
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| − | D = (b*B+c*C)/(b+c);
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| − | draw(A--B--C--A--D);
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| − | MA(B,A,D,0.5,black);
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| − | MA(D,A,C,0.6,black);
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| − | label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1));
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| − | 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));
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| − | dot(A^^B^^C^^D);
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| − | </asy>
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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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| − | <asy>
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| − | size(200);
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| − | defaultpen(fontsize(10));
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| − | real a,b,c,d,m,n;
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| − | pair A=(1,4), B=(-5,0), C=(3,0), D, E;
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| − | b = abs(C-A);c = abs(B-A);
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| − | D = (b*B+c*C)/(b+c);
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| − | m = abs(B-D);n = abs(C-D);
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| − | E = C+(B-A)*n/m;
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| − | draw(A--B--C--A--E--C);
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| − | MA(B,A,D,0.5,blue+linewidth(1));
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| − | MA(D,A,C,0.6,blue+linewidth(1));
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| − | MA(C,E,A,0.6,blue+linewidth(1));
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| − | MA(C,B,A,0.6,green+linewidth(1));
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| − | MA(B,C,E,0.6,green+linewidth(1));
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| − | label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(1,-1));label("$E$",E,(0,-1));
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| − | 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));label("$b$",(E+C)/2,(1,-0.5));
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| − | dot(A^^B^^C^^D^^E);
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| − | </asy>
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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]], with 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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| − | | |
| − | === Method 3 ===
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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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| − | <math>[ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB.</math>
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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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| − | | |
| − | <math>[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.</math>
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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>.
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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. | |
| − | # 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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