Question

Show that the sum of three altitudes of a triangle is less than the sum of the three sides of a triangle

Answer 1

Let there be a triangle ABC.
So, AF, BD and CE are three altitudes in the triangle ABC.
To prove : AB + BC + CA > AF + BD + CE
proof in Δ ABF, 
∠ AFB = 90°
Therefore the other two angles are acute angles
∴ ∠ AFB > ∠ABF
∵ AB > AF (side opposite to greater angle is longest)
Similarly BC > BD and AC > CE
Adding the three inequalities, we get
AB + BC + CA > AF + BD + CE
Hence proved

Answer 2

Let there be a triangle ABC with its altitudes D, E, and F from vertices A, B and C respectively.

The altitudes form a right angle at their corresponding bases. Also, in a right triangle the hypotenuse is the longest side. Taking the right triangles formed by the altitudes and the sides as the hypotenuse, we observe that in each triangle, the side forms the longest side, i.e,

In triangle ABD, AB is the longest side

In triangle ACF, AC is the longest side

In triangle CBE, BC is the longest side

So, adding all the three, we get that the perimeter of a triangle is greater than the sum of its three altitudes.

Hope the answer is helpful !!

Good Day !!