In ΔABC, AD bisects ∠A and intersects BC at D.  If BC = 5, AB = 4, AC = 8, then BD = ?

\(\frac{20}{11}\) \(\frac{5}{4}\) \(\frac{5}{3}\) \(\frac{20}{13}\)

\(\frac{5}{3}\)

\(\frac{BD}{DC}\) = \(\frac{4}{8}\)           [interior angle bisector theorem, AD cuts the opposite sides in the ratio of remaining sides] BD = 4R , DC = 8R, therefore BC = 12R BC =12R= 5 (given) 1R = \(\frac{5}{12}\) BD = \(\frac{5}{12}\) × 4 = \(\frac{5}{3}\)