Question

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

Answer 1

$\underline \purple{ANSWER :-}$

Let ABC be the △ with sides BC,AC and AB opposite to the angles A,B and C being in length equal to a,b and c respectively.

To prove:- AL + BM + CN < AB + BC + CA

Construction:- Draw perpendiculars AL,BM and CN from A,B and C to opposite sides meeting respectively.

Now,

In △ACL,

Using pythagoras theorem,

AC² = AL² + CL²

=> AC² > AL²

=> AC > AL .....(1)

Similarly,

AB > BM .....(2)

BC > CN .....(3)

Adding equation (1), (2) & (3), we get

AB + BC + AC > AL + BM + CN

Hence proved.

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.