Is there a triangle $ABC$ having following properties? :
$ABC$ is not right angled
$d(A,B)\ne d(A,C)$
If one draws the largest possible square such that one of the side of square is a subset of $AB$ and rest two vertexes are on sides $AC$ and $BC$ and similarly if one draws the largest possible square such that one of the side of square is a subset of $AC$ and rest two vertexes are on sides $AB$ and $BC$, the areas of squares are the same.
No such triangle exists. Suppose the contrary, that it did. Take a look at the picture.
Triangles $ATU$ and $APS$ are equal, since they have a common angle at $A$, they have right angles at $T$ and $P$ respectively, and side $TU$ is equal to side $PS$ (since if the two squares have the same area, then their sides are equal too). So $AP$ equals $AT$, and if $X$ is the intersection point of $TU$ and $PS$ then triangles $APX$ and $ATX$ are equal. Hence the angular bisector $b$ at $A$ goes through $X$, and it easily follows that all other element in the picture are symmetric about $b$, in particular $R$ is symmetric to $V$, hence the line $RV$ is perpendicular to $b$, and $B$ and $C$ are symmetric, so $AB$ is equal to $AC$.
It looks like there may be a gap in the above proof: What if $R$ and $V$ coincide? Then $R$ and $V$ would not determine a unique line, and $B$ and $C$ need not be symmetric! But, if $R$ and $V$ did coincide, then $W$ and $S$ would also have to coincide, for otherwise, as we have a right angle at $W$, the hypotenuse $RS$ of triangle $RWS$ would have to be longer than the side $RW$, but they are equal as the sides of squares of equal area. And, as $W$ ans $S$ must also coincide, then the two squares coincide, in particular points $T$, $P$ and $A$ all coincide, as $AB$ would have to be perpendicular to $AC$.
As Irvan has remarked in his comment, the triangle $\triangle$ has to be acute in order to make both envisaged squares possible.
Consider the square standing on the side $b=AC$, and let $h_b$ be the height from $B$ to $b$. The side length $s$ of the square has to satisfy $${s\over h_b-s}={b\over h_b}\ ,$$ from which we obtain $$s={b\>h_b\over b+h_b}\ .$$ Similarly, the square standing on the side $c=AB$ has side length $$s'={c\>h_c\over c+h_c}\ .$$ Now a priori $$b\>h_b=c\>h_c=2{\rm area}(\triangle)\ .\tag{1}$$ Therefore the condition $s'=s$ enforces $$c+h_c=b+h_b\quad (=:\ell)\ .\tag{2}$$ Now comes the tricky part: From $(1)$ and $(2)$ it follows by Vieta's theorem that both two-element sets $\{b, h_b\}$ and $\{c, h_c\}$ solve the "artificial" quadratic equation $$p^2- \ell\> p+2{\rm area}(\triangle)=0$$ in the auxiliary variable $p$. This implies $\{b, h_b\}=\{c, h_c\}$ (as sets). Now $b\ne c$ by assumption; furthermore $h_b<c$, since $\triangle$ is acute. This shows that a configuration of the envisaged kind is impossible.
