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Question
Prove that the sum of three altitudes of a triangle is less than the sum of its sides.
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Solution
We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.
In ΔABC we have
AD ⊥BC,BE ⊥ AC and CF ⊥ AB
We have to prove
AD BE + CF < AB + BC + AC
As we know perpendicular line segment is shortest in length
Since AD ⊥ BC
So AB >AD ........(1)
And
AC > AD ........(2)
Adding (1) and (2) we get
AB + AC > AD + AD
AB + AC > 2AD ........(3)
Now BE ⊥ AC, so
BC + BA > BE + BE
BC + BA > 2BE .......(4)
And againCF ⊥ AB , this implies that
AC + BC > 2AF ........(5)
Adding (3) & (4) and (5) we have
(AB + AC ) + (AB + BC )+ (AC + BC) >2AD + 2BE + 2CF
⇒ 2 (AB + BC + AC)>2(AD + BE + CF)
Hence AD BE + CF < AB + BC + AC Proved.
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