In the given figure, P is a point on the side BC of triangle ABC. Prove that (AB + BC +AC)>2AP.
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Answers
Answered by
56
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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Answered by
21
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
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