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प्रश्न
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
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उत्तर
We have to prove that the perimeter of a triangle is greater than the sum of its altitude.
In ΔABC
AD⊥ BC , BE ⊥ AC , CF⊥AB
We have to prove
AB + BC + CD > AD + BE + CF
Since AD⊥ BC
So AB > AD and AC > AD
By adding AB + AC > AD + AD, we have
AB + AC > 2AD ........(1)
Now consider BE ⊥ AC then
BC > BE, and BA > BE
Now by adding BC + BA > 2BE .......(2)
Again consider CF⊥AB
AC > CF, and BC > CF
By adding AC + BC > 2CF ...........(3)
Adding (1), (2) and (3), we get
2(AB + BC + CA)>2 (AD + BE + CF)
⇒ AB + BC + CA > AD + BE + CF
Hence the perimeter of a triangle is greater than the sum of all its altitude.
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| 1 | PM = QM | 1 | ... |
| 2 | ∠PMA = ∠QMA | 2 | ... |
| 3 | AM = AM | 3 | ... |
| 4 | ΔAMP ≅ ΔAMQ | 4 | ... |
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