Question

ABC is a triangle. AD is the bisector of Angle BAC. Prove that AB is greater than BD.

Answer 1

AD is the internal bisector of <A. Using standard nomenclature, c/b =BD/DC by the angle bisector theorem or c/(c+b)=BD/(BD+DC)=BD/BC= BD/a or BD =ac/(b+c). Now by the triangle inequality, b+c>a and therefore 1>a/(b+c) or c>ac/(b+c) or AB> BD