3) In the figure below, ABC is a triangle and X is any
point on BC. Show that AB + AC + BC > 2AX
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3) In the figure below, ABC is a triangle and X is any
point on BC. Show that AB + AC + BC > 2AX
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AB+BX>AX....eq(1)
AC+CX>AX....eq(2),
adding these two equations, we get
AB+BC+AC>2AX
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Verified answer
[tex]\huge \pink{Aŋʂῳɛཞ\implies}[/tex]
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:
[tex]\begin{gathered}\bigtriangleup ABX \\AB + BX > AX..(1)\end{gathered} \\
△ABX \\
AB+BX>AX..(1)[/tex]
"The sum of any two sides of any triangle is always greater than third side"
similarly in
[tex]\begin{gathered}\bigtriangleup AXC \\AC + CX > AX..(2)\end{gathered}[/tex]
[tex]△AXC
AC+CX>AX..(2)[/tex]
adding 1 and 2
we get,
AB+BX+AC+CX > AX+AX
(BX+CX = BC)
Therefore,
AB +AC+BC >2AX
hence proved .
[tex]\mathcal\purple{HOPE \: ITS \: HELP \: YOU}[/tex]