Question

In the adjoining figure,P is a point on the side BC of∆ABC . prove that
(AB+BC+CA) >2AP
​

Answer 1

Answer:

Hey I hope I solved your problem

Step-by-step explanation:

In triangle ABP,

(Sum of any two sides is greater than 3rd side)

AB+BP>AP (1)

Similarly, in triangle ACP,

(Sum of any two sides is greater than 3rd side)

AC+CP>AP. (2)

By adding 1 and 2 We have,

AB+BP+AC+CP>AP+AP

AB+BC+AC>2AP. ( Since, BP+CP=BC )