In ΔABC, AB = 7 cm, BC = 10 cm, and AC = 8 cm. If AD is the angle bisector of ∠BAC, where D is a point on BC, then DC (in cm) = ?

$\frac{16}{3}$

As we know,

If AD bisect \(\angle\)BAC, then,

⇒ \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)

⇒ \(\frac{7}{8}\) = \(\frac{BD}{DC}\)

⇒ DC = \(\frac{8BD}{7}\)  ..(1.)

⇒ BD + DC = BC

⇒ BD + \(\frac{8BD}{7}\) = 10

⇒ \(\frac{15BD}{7}\) = 10

⇒ BD = 10 x \(\frac{7}{15}\)

⇒ BD = \(\frac{14}{3}\)

Putting in equation (1.)

⇒ DC = \(\frac{14}{3}\) x \(\frac{8}{7}\)

⇒ DC = \(\frac{16}{3}\)