Question

Prove that the sum of three altitudes of a triangle is less than the sum of its sides.

Answer 1

We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.

In ΔABC we have

 AD ⊥BC,BE ⊥ AC  and  CF ⊥ AB

We have to prove 

  AD BE + CF < AB + BC + AC

As we know perpendicular line segment is shortest in length

Since  AD ⊥ BC

So  AB >AD    ........(1)

And 

AC > AD       ........(2)

Adding (1) and (2) we get

 AB + AC > AD + AD

  AB + AC > 2AD  ........(3)

Now BE ⊥ AC, so

BC + BA > BE + BE

 BC + BA > 2BE     .......(4)

And againCF ⊥ AB , this implies that

 AC + BC > 2AF   ........(5)

Adding (3) & (4) and (5) we have

(AB + AC ) + (AB + BC )+ (AC + BC) >2AD + 2BE + 2CF

⇒ 2 (AB + BC + AC)>2(AD + BE + CF)

Hence AD BE + CF < AB + BC + AC Proved.