Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Given, ΔABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.
AD + BE + CF < AB + BC + AC
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.
AD ⊥ BC
AB > AD and AC > AD
AB + AC > 2AD .... (i)
BE ⊥ AC
BA > BE and BC > BE
BA + BC > 2BE ... (ii)
CF ⊥ 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)
AB + BC + CA > AD + BE + CF
The perimeter of the triangle is greater than that the sum of its altitudes
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