Question

prove that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle​

Answer 1

Prove that sum of three altitude of a triangle less than the sum of the three sides of a triangle

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

THANKS!!

Answer 2

Hi

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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

hope it helped!

Thanks!!