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Then \[AB + AD > BC\]

If the above statement is true, then mention answer as 1, else mention 0 if it is false

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According to this rule, In\[\vartriangle ABD\],

\[AB + AD > BD\]

\[AB + AD > BE + ED\]--- (1)

Also, in \[\vartriangle BCE\], we have sum of two sides greater than that of its third side

\[BE + EC > BC\]-- (2)

After Adding equations \[1,2\] we get

\[AB + AD + BE + EC > BE + ED + BC\]

Subtracting \[BE\] from both sides

\[AB + AD + EC > ED + BC\]

But it is given for the two sides of the given figure, \[EC = ED\]

So, the two will be cancelled from each side, And we are only left with

Hence, \[AB + AD > BC\]