Problem:
Triangle ABC has AB=27,AC=26, and BC=25. Let I denote the intersection of the internal angle bisectors of △ABC. What is BI ?
Answer Choices:
A. 15
B. 5+26+33
C. 326
D. 32546
E. 93 Solution:
Let a=BC,b=AC, and c=AB. Let D,E, and F be the feet of the perpendiculars from I to BC,AC, and AB, respectively. Because BF and BD are common tangent segments to the incircle of △ABC, it follows that BF=BD. Similarly, CD=CE and AE=AF. Thus
Let s=21(a+b+c)=39 be the semiperimeter of △ABC and r=DI the inradius of △ABC. The area of △ABC is equal to rs and also equal to s(s−a)(s−b)(s−c) by Heron's formula. Thus
r2=s(s−a)(s−b)(s−c)=3914⋅13⋅12=56.
Finally, by the Pythagorean Theorem applied to the right triangle BDI, it follows that