Given: ΔABC in which AD⊥BC, BE⊥AC and CF⊥AB
To prove:
AD + BE + CF < AB + BC + AC
Figure:
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,
Adding (1), (2) and (3), 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