Bisector
Division of bisector
Let a triangle 
be given.
Let 
and 
be the bisectors of 
he segments 
and 
meet at point 
Find ![\[\frac {BI}{BB'}, \frac {DA'}{DC'}, \frac {BD}{BB'}.\]](http://latex.artofproblemsolving.com/5/1/4/514f96e06e7dd7f44797855eb8ce5f956663504a.png)
Solution
![\[\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.\]](http://latex.artofproblemsolving.com/4/f/d/4fde431607d6e48344816a4f464a1e111879b731.png)
Similarly 
![\[\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.\]](http://latex.artofproblemsolving.com/c/2/9/c29e89101ea63d35bfc04f787c84588fba11f410.png)
![\[\frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.\]](http://latex.artofproblemsolving.com/a/a/4/aa4500b7d5b2f65f09c84e3455da66dccb16ca1d.png)
Denote 
Bisector 
Bisector 
![\[\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.\]](http://latex.artofproblemsolving.com/3/4/e/34edec2d0145ed29a1b4137b26a852d12efdc5e3.png)
vladimir.shelomovskii@gmail.com, vvsss
Proportions for bisectors
The bisectors 
and 
of a triangle ABC with 
meet at point 
Prove 
Proof
Denote the angles 
and 
are concyclic.
![\[\angle BEA = \angle BEI = \angle ADC.\]](http://latex.artofproblemsolving.com/8/1/7/817ef0fe8e248ac346bd9211c3648f658f76cc16.png)
The area of the 
is
![\[[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies \frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.\]](http://latex.artofproblemsolving.com/a/9/d/a9d0d9aadc6b42d2706ed3d4c49d9ce59ce8ac5c.png)
![\[\frac {DI}{IE} = \frac {DI}{DC} \cdot \frac {AE}{IE}\cdot \frac {DC}{AI}= \frac {c}{a+b+c} \cdot \frac {a+b+c} {a} \cdot \frac {a}{c} = 1.\]](http://latex.artofproblemsolving.com/e/6/1/e61dd3c88cb5176bc2787f68d7224b9ced90b8fb.png)
vladimir.shelomovskii@gmail.com, vvsss
