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draw a triangle ABC and three altitudes

AH

BL

and CM

AH_|_ BC

BL _|_AC

and CM _|_ AB

for right triangle AHB, using Pythagorean

AH² = AB² - BH²

so AH < AB

similarly for other altitudes

thus

AH+BL+CM is < AB+BC+CA

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Q: Prove that the perimeter of a triangle is greater then its three altitudes?
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