Let $ABC$ be triangle with sides that are not equal. Find point $X$ on $BC$ From following conditions. Let $ABC$ be triangle with sides that are not equal. Find point $X$ on $BC$ such that $$\frac{\text{area}\ \triangle{ABX}}{\text{area}\ \triangle{ACX}}=\frac{\text{perimeter}{\ \triangle{ABX}}}{\text{perimeter}{\ \triangle {ACX}}}$$
I am not getting how to start with, Clearly  $ABC$ triangle is scalene but what about other triangle is it necessary they should be scalene too? Please help me to solve this. Thanks in advance

Obviously, the stated condition means that
$\triangle ABX$ and $\triangle AXC$
must have the same radius of the inscribed circle.
That means $AX$ must be the incircle bisector of $\triangle ABC$,
see
Incircle bisectors and related measures.



 For $A$, $B$, $C$
 ordered in positive (counterclockwise) direction,
 $|BC|=a$, $|AC|=b$, $|AB|=c$, using expressions given in the link above,
 we can find out that

 \begin{align}|BX|&=\tfrac12\,a-\frac{b-c}{2a}\,\Big(b+c-\sqrt{(b+c)^2-a^2}\Big).\end{align}


