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**Hint:**We know that the sides opposite to the equal sides are equal. Similarly the side opposite to the larger angle is larger.Using this concept we try prove the statement.

**Complete step-by-step answer:**

Generally this is the theorem that the side opposite to the larger angle is larger. So now let us see how.

Let $ABC$ be a triangle.

Let us assume that $\angle ABC > \angle ACB$

So if we prove that $AC > AB$, then it is clear that the side opposite to the larger angle is larger.

For this proof we need the construction.

So construct $BD$ on $AC$ such that $\angle DBC = \angle ACB$

As $\angle ABC > \angle ACB$, so we can find a point on $AC$ that is $D$ such that $\angle DBC = \angle ACB$.

As construction is done, it divides $\Delta ABC$ into two triangles that are $\Delta ABD$ and $\Delta DBC$.

Now in $\Delta DBC$,

$\angle DBC = \angle ACB$

As we constructed $BD$ such that the two angles are equal.

Now we know that in the triangle, the sides opposite to equal angles are equal.

So $BD = DC$ $ - - - - - - - \left( 1 \right)$

Now we can add $AD$ on both sides of this above equation,

$AD + BD = DC + AD$

Now if solve $AD + DC$, we get $AC$

So $AD + BD = AC$

Or we can say that,

$AC = BD + AD$ $ - - - - - - - \left( 2 \right)$

Now in $\Delta ADB$,

We know that for any triangle to form, the sum of two sides will always be greater than the third side.

So $BD + AD > AB$ $ - - - - - - - - \left( 3 \right)$

Now putting the value of $AD + BD = AC$ in equation (3),

So we get $AC > AB$.

Hence it is proved that $AC > AB$ if $\angle ABC > \angle ACB$

Hence we can say that the side opposite to the larger angle is larger.

**Note:**Let for any triangle $ABC$, by applying the $\sin $ rule we get,

$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$

So if $\angle A > \angle B$,

$\sin A > \sin B$ as $\sin $ is the increasing function.

So $\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b}$

$a = \dfrac{{\sin A}}{{\sin B}}b$ and $\sin A > \sin B$

So $a > b$

Hence Proved.

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