In triangle ABC, AD is the bisector of ∠A. If AB = 5 cm, AC = 7.5 cm and BC = 10 cm, then what is the distance of D from the mid-point of BC (in cm) ?

1

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

= \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)  [By angle bisector theorem]

= \(\frac{5}{7.5}\) = \(\frac{BD}{DC}\)

= \(\frac{2}{3}\) = \(\frac{BD}{DC}\)

= \(\frac{3}{2}\) = \(\frac{DC}{BD}\)

Adding 1 to both sides, we get

= \(\frac{3}{2}\) + 1 = \(\frac{DC}{BD}\) + 1

= \(\frac{5}{2}\) = \(\frac{BD\;+\;DC}{BD}\)= \(\frac{BC}{BD}\)

Since BC = 10, we get

The value of BD = (\(\frac{2}{5}\) x 10) = 4

The value of DC = (\(\frac{3}{5}\) x 10) = 6

ACT

Midpoint of BC = \(\frac{10}{2}\) = 5

So, we take M as midpoint on BC = 5

Now,

The distance between D and M = (6 - 5) = 1

Therefore, answer 1 cm.