Prove tht sum of 3 altitudes of a triangle is less than the sum of three sides of a triangle

The shortest distance from a point to a line is the perpendicular. Hence each altitude is shorter than both of the triangles sides that that are incident at the angle the altitude is from. I.e. Altitude from A is shorter than both b and c, altitude from B is shorter than both both a and c, the altitude from C is shorter than both a and b. Now adding selectively the result follows. Note: when the triangle is a right triangle one, but only one, of the inequalities is an equality. So the result is still a strict inequality.

In triangle APB, AP is a median.

Angle APB is greater than angle ABP. so angles opposite to greater side is longer.

therefore AB is greater than AP. hence proved that in triangle APB the side of the triangle is more than the median or altitude AP of the triangle.

In the same manner the other triangles can also be proved.

I am not describing the whole answer because practice makes a man perfect!!!

Its very easy. It is just in different triangles.

Hope it helps!!!

cheeeerzzz!!!

Now the other triangles you have to check are triangles APC,ACM,ABQ.

Now please solve it!!!!

Hope it helps you again!!!!