$A B C$ is an equilateral triangle and $D$ is any point in $A C$ . Prove that $B D > A D$ .

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Solution

Consider the triangle $△ A D B$ .
$∠ B A D = 60^{\circ}$ and $∠ A B D < ∠ A B C ⟹ ∠ A B D < 60^{\circ}$

Now, in a triangle, side opposite larger angle is greater than the side opposite smaller angle.
Hence, as BD is opposite $∠ B A D$ , AD is opposite $∠ A B D$ and we saw that $∠ A B D < ∠ B A D = 60^{\circ},$

$∴ B D > A D$

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