SHOWTHAT THE THREE ALTITUDES OF A TRIANGLE IS LESS THAN THE SUM OF THE THREE SIDES OF A TRIANGLE
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Take triangle ABC in which AD,BE,CF are the medians to the sides BC,AC, and AB respectively.
We know that, the sum of any two sides of triangle is greater than twice the median bisecting the third side
AB+AC>2AD.....(1)
AB+BC>2BE.......(2)
BC+AC>2CF.....(3)
Adding eq. 1. 2 an 3, we get-
AB+AC+AB+BC+BC+AC>2AD+2BE+2CF
=2AB+2BC+2AC>2AD+2BE+2CF
=2(AB+BC+AC)>2(AD+BE+CF)
AB+BC+AC>AD+BE+CF
Proved ....hope I helped you...!!
We know that, the sum of any two sides of triangle is greater than twice the median bisecting the third side
AB+AC>2AD.....(1)
AB+BC>2BE.......(2)
BC+AC>2CF.....(3)
Adding eq. 1. 2 an 3, we get-
AB+AC+AB+BC+BC+AC>2AD+2BE+2CF
=2AB+2BC+2AC>2AD+2BE+2CF
=2(AB+BC+AC)>2(AD+BE+CF)
AB+BC+AC>AD+BE+CF
Proved ....hope I helped you...!!
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