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:
Answer & explanation
Correct answer: option 2
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, \(\frac{DC}{4}\) = \(\frac{4}{3}\)