In ΔABC, AB = 7 cm. BC = 10 cm, AC = 8 cm. If AD is the angle bisector of ∠BAC, where D is a point on BC, then $\frac{DC}{4}$(in cm) is equal to:

$\frac{4}{3}$

AD is the angle bisector of \(\angle\)BAC

So, \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)

⇒ \(\frac{7}{8}\) = \(\frac{BD}{DC}\)

⇒ 88D = 7DC

⇒ BD = \(\frac{7DC}{8}\)     ..(1)

 Here, BD + DC = BC

⇒ \(\frac{7DC}{8}\) + DC = BC = 10 [As, BD = \(\frac{7DC}{8}\)]

⇒ \(\frac{15DC}{8}\) = 10

⇒ DC = 10 x \(\frac{8}{15}\) = \(\frac{80}{15}\) = \(\frac{16}{3}\)

Therefore, \(\frac{DC}{4}\) = \(\frac{4}{3}\)