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