Difference between revisions of "Angle Bisector Theorem"
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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> | <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> | ||
− | == | + | == 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. | |
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− | <asy> | + | <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> |
− | size(200); | ||
− | defaultpen(fontsize( | ||
− | real a,b,c,d | ||
− | pair A=(1, | ||
− | b = abs(C-A);c = abs(B-A); | ||
− | D = (b*B+c*C)/(b+c); | ||
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− | draw(A--B--C--A-- | ||
− | MA(B,A,D, | ||
− | MA(D,A,C, | ||
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− | label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,( | ||
− | 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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− | </asy> | ||
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== Examples == | == Examples == |
Revision as of 23:56, 17 August 2016
This is an AoPSWiki Word of the Week for June 6-12 |
== Introduction ==read
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then
. Likewise, the converse of this theorem holds as well.
Introduction
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then
. Likewise, the converse of this theorem holds as well.
Examples
- 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.