Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Explanation

Given: `triangleABC` in which `ADbotBC, BEbotAC` and `CFbotAB`
To prove: `AD + BE + CF < AB + BC + AC`
Proof:-We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e., the perpendicular line segment is the shortest.
Therefore,
`ABbotBC`
`AB>AD` and `AC>AD`
`AB+AC>2AD............(I)`
Similarly `BEbotAC`
`BA>BE` and `BC>BE`
`BA+BC>2BE............(II)`And also `CFbotAB`
`CA>CF` and `CB>CF`
`CA+CB>2CF.......(III)`
Adding (I), (II) and (III), we get
`AB+AC+BA+BC+CA+CB>2AD+2BE+2CF`
`2AB+2BC+2CA>2(AD+BE+CF) `
`2(AB+BC+CA)>2(AD+BE+CF)`
`AB+BC+CA>AD+BE+CF`
The perimeter of the triangle is greater than the sum of its altitudes Therefore, Hence proved