Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle.
ΔABC, in which altitudes AQ, BR and CP meet at M.
To Prove:
AB + BC + CA > BR + CP + AQ
I Proof:
In ΔABQ, we have ∠AQB = 90°
... ∠ABQ will be a acute angle
...∠AQB > ∠ABQ
...AB > AQ ...... (i)
Similarly, we can prove that:
BC > BR ...... (ii)
and CA > CP ...... (iii)
Adding three inequalities, we get
AB + BC + CA > AQ + BR + CP
II Proof:
AB > AQ, AC > AQ ...... [ ... Perpendicular is shortest]
... AB + AC > 2AQ ...... (i)
Similarly BC + CA > 2PC ...... (ii)
and AB + BC > 2BR
Adding (i), (ii) and (iii), we get
2AB + 2BC + 2CA > 2AQ + 2BR + 2CP
or AB + BC + CA > AQ + BR + CP
To Prove:
AB + BC + CA > BR + CP + AQ
I Proof:
In ΔABQ, we have ∠AQB = 90°
... ∠ABQ will be a acute angle
...∠AQB > ∠ABQ
...AB > AQ ...... (i)
Similarly, we can prove that:
BC > BR ...... (ii)
and CA > CP ...... (iii)
Adding three inequalities, we get
AB + BC + CA > AQ + BR + CP
II Proof:
AB > AQ, AC > AQ ...... [ ... Perpendicular is shortest]
... AB + AC > 2AQ ...... (i)
Similarly BC + CA > 2PC ...... (ii)
and AB + BC > 2BR
Adding (i), (ii) and (iii), we get
2AB + 2BC + 2CA > 2AQ + 2BR + 2CP
or AB + BC + CA > AQ + BR + CP
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