Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Answers
The perimeter of a triangle is greater than the sum of its altitudes.
Step-by-step explanation:
In triangle ABC, AG bisects BC, CF bisects AB and BE bisects CA.
Proof :
Perimeter of a ΔABC = AB+BC+AC
Altitudes of ΔABC are AG, BE, CF
To prove: (AB+BC+AC) > (AG+BE+CF)
In a triangle the sum of any two sides is greater than twice the median bisecting the third side.
In triangle ABC, AG is the median bisecting BC
AB+AC >2 AG …………. (1)
Similarly, CF is the median bisecting AB
BC+CA >2 CF ………….. (2)
Similarly, BE is the median bisecting AC
BC+AB >2 BE …………. (3)
Adding (1), (2), (3), we get,
= AB+AC+BC+AC+BC+AB >2AG+2CF+2BE
= 2(AB+BC+AC) >2 (AG+CF+BE)
= AB+BC+AC > AG+BE+CF
Hence Proved.
To learn more:
1. Prove that the sum of any two sides of a triangle is greater than twice the length of median drawn to the third side
brainly.in/question/889583
2. Prove that the perimeter of triangle is greater than twice the median drawn on any side
brainly.in/question/2528441
