USA
Prove that the sum of the lengths of the three altitudes of a triangle is less than the sum of the lengths of the three sides of the triangle.
Answer:
- Let ^@AP,^@ ^@BQ,^@ and ^@CR^@ be the altitudes of ^@\triangle ABC.^@
- We know that the perpendicular is the shortest line segment that can be drawn from a point outside the line to that line.
Thus, we have @^ \begin{aligned} AP < AB && [\text { As AP is the perpendicular from the point A.}] \\ BQ < BC && [\text { As BQ is the perpendicular from the point B.}] \\ CR < AC && [\text { As CR is the perpendicular from the point C.}] \\ \end {aligned} @^ Adding the three equations, we have @^ \begin{aligned} & AP + BQ + CR < AB + BC + AC \\ \implies & \text{ Sum of the lengths of the altitudes } < \text{ Sum of the lengths of the sides } \end{aligned}@^ - Thus, the sum of the lengths of the three altitudes of a triangle is less than the sum of the lengths of the three sides of the triangle.
