Prove that the sum of the three altitudes of a triangle is less than the sum of three sides of a triangle.
Answers
Answer:
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
Answer:
ABC is a triangle ,AD ,BE and CF are Altitudes on side
BC ,AC and AB
Now in ABD , D is 90° ,B is acute angle
So D > B
AB > AD -------(1) [side opposite to the greater angle is also greater]
And, In ADC , D is 90° ,C is acute angle
So D > C
AC > AD -------(2) [side opposite to the greater angle is also greater]
And, In BFC , F is 90° ,B is acute angleSo F > B
BC > CF -------(3) [side opposite to the greater angle is also greater]
And, In AFC , F is 90° ,A is acute angle
So F > A
AC > CF -------(4) [side opposite to the greater angle is also greater]
And, In ABE , E is 90° ,A is acute angle
So E > A
AB > BE -------(5) [side opposite to the greater angle is also greater]
And, In BEC , E is 90° ,C is acute angle
So E > C
BC > BE -------(6) [side opposite to the greater angle is also greater]
On Adding eq (1)(2)(3)(4)(5)and (6)
AB+AC+BC+AC+AB+BC > AD+AD+CF+CF+BE+BE
2AB+2BC+2AC > 2AD +2CF+2BE
2(AB+BC+AC) > 2(AD+CF+BE)
AB+BC+AC > AD +BE +CF
Hence Proved.