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Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of a triangle.

sorry there is no figure for this question.. the figure attached is wrong .plz don't concentrate.



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Please keep drawing the figure as you follow below.
Let ABC be the triangle with sides BC, AC, AB opposite angles A, B, C being in length equal to a, b, c respectively.

Draw perpendiculars AD, BE and CF from A, B, C to opposite sides meeting sides BC, AC and AB at D, E, F respectively.

Now perpendicular AD² = AC² - CD² ==> AD² < AC² or  AD < AC
                                                                           or AD < b -----(1)
                  Also,  BE² = AB² - AE² ==> BE² < AB² or BE < AB
                                                                           or BE < c ------(2)
            Likewise , CF² = BC² - BF² ==> CF² < BC² or CF < BC 
                                                                           or CF < a ----(3)
Adding inequalities (1), (2), (3), AD + BE + CF < a + b + c
10 3 10
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