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Question

Show that the sum of the three altitudes of a triangle is less than the sum of three sides of the triangle. [4 MARKS]


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Solution

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

Given: A ABC in which ADBC, BEAC and CFAB.

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 line segment is the shortest.

Therefore,

AD BC AB > AD and AC > AD

AB+AC > AD + AD

AB+AC > 2 AD .....(i)

BE AC BC > BE and BA > BE

BC + BA> BE + BE

BA + BC > 2 BE ...(ii)


And, CF AB AC > CF and BC > CF

AC + BC > 2 CF ...(iii)


Adding(i),(ii) and (iii), we get

(AB + AC) + (AB + BC) + (AC + BC) > 2 AD + 2 BE + 2 CF

2(AB + BC + AC) > 2(AD + BE +CF)

AD + BE + CF < AB + BC + AC


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