|
|
| (26 intermediate revisions by 12 users not shown) |
| Line 1: |
Line 1: |
| − | {{WotWAnnounce|week=June 6-12}}
| + | #REDIRECT[[Angle bisector theorem]] |
| − | | |
| − | == Introduction ==
| |
| − | 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,red);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 ==
| |
| − | ===Method 1 ===
| |
| − | 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:
| |
| − | | |
| − | <asy>
| |
| − | size(200);
| |
| − | defaultpen(fontsize(10));
| |
| − | real a,b,c,d,m,n;
| |
| − | pair A=(1,4), B=(-5,0), C=(3,0), D, E;
| |
| − | b = abs(C-A);c = abs(B-A);
| |
| − | D = (b*B+c*C)/(b+c);
| |
| − | m = abs(B-D);n = abs(C-D);
| |
| − | E = C+(B-A)*n/m;
| |
| − | draw(A--B--C--A--E--C);
| |
| − | MA(B,A,D,0.5,blue+linewidth(1));
| |
| − | MA(D,A,C,0.6,blue+linewidth(1));
| |
| − | MA(C,E,A,0.6,blue+linewidth(1));
| |
| − | MA(C,B,A,0.6,green+linewidth(1));
| |
| − | MA(B,C,E,0.6,green+linewidth(1));
| |
| − | 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));
| |
| − | 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));
| |
| − | dot(A^^B^^C^^D^^E);
| |
| − | </asy>
| |
| − | | |
| − | 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.
| |
| − | | |
| − | By [[AA similarity]], <math> \triangle DAB \cong \triangle DEC </math>. By the properties of similar triangles, we arrive at our desired result:
| |
| − | | |
| − | <center><math> \frac cm = \frac bn.</math> </center>
| |
| − | | |
| − | === Method 2 ===
| |
| − | Since B,D, and C are collinear, <math>\dfrac{[ABD]}{[ADC]} = \dfrac{m}{n}</math>. Now consider the perpendicular from <math>D</math> to <math>AB</math> and <math>D</math> to <math>AC</math>. Every point on the angle bisector of an angle is equidistant to the sides of the angle, so the height <math>h</math> to <math>AB</math> is equal to the height to <math>AC</math>. Thus <math>\dfrac{[ABD]}{[ADC]} = \dfrac{\dfrac{ch}{2}}{\dfrac{bh}{2}} = \dfrac{c}{b}</math>. Thus <math>\dfrac{c}{b} = \dfrac{m}{n}</math>, so <math>\dfrac{c}{m} = \dfrac{b}{n}</math>. We can prove the converse by the Phantom Point Method, since we can find <math>m</math> and <math>n</math> in terms of <math>a</math>, <math>b</math>, and <math>c</math>, and prove that the points are the same.
| |
| − | | |
| − | === Method 3 ===
| |
| − | Let <math>AD = d</math>. Now, we can express the area of triangle ABD in two ways:
| |
| − | | |
| − | <math>[ABD] = \frac 12 cd\sin \angle BAD = \frac 12 md \sin \angle ADB.</math>
| |
| − | Thus, <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm</math>.
| |
| − | | |
| − | Likewise, triangle ACD can be expressed in two different ways:
| |
| − | | |
| − | <math>[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.</math>
| |
| − | Thus, <math>\frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn</math>.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | We conclude that <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn</math>, 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 <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.
| |
| − | # Part '''(b)''', [[1959 IMO Problems/Problem 5]]. | |
| − | | |
| − | == See also ==
| |
| − | * [[Angle bisector]]
| |
| − | * [[Geometry]]
| |
| − | * [[Stewart's Theorem]]
| |
| − | | |
| − | [[Category:Geometry]]
| |
| − | | |
| − | [[Category:Theorems]]
| |
