Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Given: A in which AD perpendicular BC and BE perpendicular AC and CF perpendicular AB.
To prove: AD + BE + CF < AB + BC + AC
Proof: We know that all the segments that can be drawn into a given line, from a point not lying on it, perpendicular distance i.e. the perpendicular line segment is the shortest. Therefore,
AD perpendicular BC
AB > AD and AC > AD
AB + AC > 2AD (i)
Similarly,
BE perpendicular AC
BA > BE and BC > BE
BA + BC > 2BE (ii)
And also
CF perpendicular AB
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.
Hence, proved