Three sides of triangle ABC are a, b and c. a = 4700 cm, b = 4935 cm and c = 6815 cm. The internal bisector of ∠A meets BC at P and the bisector passes through incentre O. PO : OA = ?

a : b : c = 4700 : 4935 : 6815 = 20:21:29 and 20²+21² = 400 + 441 = 841 = 29² ∴ ∠ACB = 90° $\therefore \quad$ In-radius $=\frac{\mathrm{AC}+\mathrm{BC}-\mathrm{AB}}{2}$ $=\frac{20+21-29}{2}=6 \mathrm{~cm}$

∴ MO = MC = 6 cm.

∴ AM = 21 – 6 = 15 cm.

$\therefore \quad \frac{\mathrm{AM}}{\mathrm{MC}}=\frac{\mathrm{AO}}{\mathrm{OP}}=\frac{15}{6}=\frac{5}{2}$

$\therefore \quad \frac{\mathrm{PO}}{\mathrm{AO}}=\frac{2}{5}$