Question

in the figure below, ABC is a triangle and X is any
point on BC Show that AB + AC + BC > 2AX​

Answer 1

Answer:

proved.

Step-by-step explanation:

we proved it on the basis of a theorem

"The sum of any two sides of any triangle is always greater than third side"

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Answer 2

AB +AC+BC >2AX

Step-by-step explanation:

here,

given that: ABC is a triangle and  X is any  point on BC

to prove :AB + AC + BC > 2AX​

proof: in

$\bigtriangleup ABX \\AB + BX >AX..(1)$

"The sum of any two sides of any triangle is always greater than third side"

similarly  in

$\bigtriangleup AXC \\AC + CX >AX..(2)$

adding 1 and 2

we get,

AB+BX+AC+CX > AX+AX

(BX+CX = BC)

Therefore,

AB +AC+BC >2AX

hence  proved .

#Learn more:

ABC is a triangle, a circle touches sides AB AC produced and BC at X. Y and z resp  .Show that perimeter ABC = 2AX.​

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