Question

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

Answer 1

AB+BX>AX....eq(1)

AC+CX>AX....eq(2),

adding these two equations, we get

AB+BC+AC>2AX

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Answer 2

$\huge \pink{Aŋʂῳɛཞ\implies}$

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:

$\begin{gathered}\bigtriangleup ABX \\AB + BX > AX..(1)\end{gathered} \\ </p><p>△ABX \\ </p><p>AB+BX>AX..(1)$

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

similarly in

$\begin{gathered}\bigtriangleup AXC \\AC + CX > AX..(2)\end{gathered}$

$△AXC</p><p>AC+CX>AX..(2)$

adding 1 and 2

we get,

AB+BX+AC+CX > AX+AX

(BX+CX = BC)

Therefore,

AB +AC+BC >2AX

hence proved .

$\mathcal\purple{HOPE \: ITS \: HELP \: YOU}$