prove that perimeter of triangle is greater than sum of its altitude
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Given: ΔABC in which AD⊥BC, BE⊥AC and CF⊥AB
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, 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
therefore, the perimeter of the triangle is greater than the sum of its altitudes.
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