Question

the sum of three sides of a triangle is more than the sum of its altitude it helps me please

Answer 1

Answer:

The key is to use the fact that the shortest way from a point to a line is going "straight" to the line and at right angles to it.

Call your triangle ABC and let D, E, F be the feet of the altitudes from A, B, C, respectively.

Then AD is the shortest route from A to BC, so AD ≤ AB (with equality only if B=D, so ABC has a right angle at B).

Similarly, BE ≤ BC (with equality only if ABC has a right angle at C), and CF ≤ CA (with equality only if ABC has a right angle at A).

Adding these three inequalities, we have

AD + BE + CF ≤ AB + BC + CA.

But further, since the triangle ABC cannot have right angles at all three vertices, not all of the three inequalities can actually be "=" (i.e. it must be strictly "<" somewhere), so we actually have

AD + BE + CF < AB + BC + CA.

That is, the sum of the alititudes is less than the sum of the sides.

Answer 2

Of all the line segments drawn to a given line, from a point not on the line, the perpendicular is the shortest,   

                 

AL < AB, BM < BC and CN < CA


Adding the above inequalities,


AL + BM + CN < AB + BC + CA


Hence, the sum of the altitudes is less than the perimeter of the triangle.

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