prove that The sum of Three altitude of triangle is less than the sum of three side of angles
Answers
Let the altitudes through A, B, C be AD, BE, CF
AB > AD (Hypotenuse is longer than any side of the right triangle [ABD])
Similarly, AC > AD
Adding both, AB + AC > 2AD
Similarly, AB + BC > 2 BE & BC + AC > 2CF
Adding all three eqns.
2(AB + BC + AC) > 2(AD + BE + CF)
AB + BC + AC > AD + BE + CF
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.