Point X belongs to the interior of a triangle ABC. The distance from this point to lines AB, BC, AC is k, l, m respectively and r is the length of an inradius of this triangle. Prove that:
Thanks in advance!
Okay maybe I'm crazy but does this prove it? :
If point X is at the centre of the incircle then the distances from this point to each line are the shortest possible and are also equal to r. Therefore, k + l + m = 3r ..... therefore k + l + m > 2r
Somebody tell me if this is right please, and if it is not, please tell me why. Thanks.
It does cover the general case because if the point is at the centre of the incircle, it means that if it were anywhere else, the sum of each length would be greater than the sum of each length from the cente, therefore it would definately be greater than 2r at any other point in the triangle.
At the incentre, the triangle area can be expressed as the sum of the areas of 3 internal triangles
At the point, the triangle area can be expressed as the sum of the areas of 3 other internal triangles
Then
Since the length of any side of a triangle is less than the sum of the other two sides, then
Therefore the inequality is true