Question

In the given figure, P is a point on the side BC of triangle ABC. Prove that (AB + BC +AC)>2AP.​

Answer 1

Step-by-step explanation:

In ∆ABP

AB+BP>AP.......(1)

In ∆ACP

AC+CP>AP.......(2)

Adding 1 and 2

AB+AC+BP+CP>2AP

AB+AC+BC>2AP

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Answer 2

Step-by-step explanation:

In triangle ABP

AB+ BP > AP…….(1)

In triangle ACP

AC + CP > AP…….(2)

By adding the equation 1 & 2

(AB+ BP > AP)+(AC + CP > AP)

we get,

AB + AC +( BP+CP) > 2AP

where BP +CP = BC ( according to figure)

Hence , AB + BC + CA > 2 AP

HENCE PROVED