In the given figure, AP bisects ∠BAC. If AB = 4 cm, AC = 6 cm and BP = 3 cm, then the length of CP is: |
3 cm 7 cm 5 cm 4.5 cm |
4.5 cm |
We are given that :- AP bisects ∠BAC ⇒ \(\frac{AB}{AC}\) = \(\frac{BP}{CP}\) \(\frac{4}{6}\) = \(\frac{3}{CP}\) CP = 4.5 cm |