Answer :

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


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