Q. 154.4( 9 Votes )

Prove that the pe

Answer :

Given: ∆ABC 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.


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,


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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