Question

Prove that the sum of 3 altitude of a triangle is less than the sum of three sides of a triangle

Answer 1

The shortest distance from a point to a line is the perpendicular. Hence each altitude is shorter than both of the triangles sides that that are incident at the angle the altitude is from. I.e. Altitude from A is shorter than both b and c, altitude from B is shorter than both both a and c, the altitude from C is shorter than both a and b. Now adding selectively the result follows. Note: when the triangle is a right triangle one, but only one, of the inequalities is an equality. So the result is still a strict inequality.
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Answer 2

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.