The question is :-
The bisector of each angle of a triangle intersects the opposite side at a point equidistant from the midpoints of other sides of the triangle. Find all such triangles.
According to the question in $\Delta ABC$ I marked the midpoints as $A',B',C'.$ therefore let the bisector from $A$ be $AX$ hence $C'X =B'X$ therefore we need to find in which triangle does that happen.But I am confused how to move further to find which type of triangle is that.
Please help out to solve this question .It is an olympiad book question.
Let $AB=2c$, $BC=2a$ and $AC=2b$ be the sides of a triangle, $D$ and $E$ the midpoints of $BC$ and $AC$. Point $M$ on side $AB$, equidistant from $D$ and $E$, lies on the perpendicular bisector of $DE$, hence $HM={1\over2}DE={1\over2}c$. Moreover: $$ BH=a\cos(\angle ABC)={a^2+c^2-b^2\over2c}={c\over2}+{a^2-b^2\over2c} $$ and $$ BM=BH+HM={c}+{a^2-b^2\over2c},\quad AM=2c-BM={c}-{a^2-b^2\over2c}. $$ Point $M$ lies on the bisector of $\angle C$ if $BM:AM=BC:AC$, that is if: $$ (a-b)\big(2c^2-(a+b)^2\big)=0. $$ This is satisfied if either $a=b$ (isosceles triangle) or $(a+b)^2=2c^2$. Analogous equations hold for the other sides of the triangle.
If the triangle we are looking for is not isosceles, its sides must then satisfy the system $$ \cases{ (a+b)^2=2c^2\\ (b+c)^2=2a^2\\ (c+a)^2=2b^2\\ } $$ which has no positive solution, because summing up the three equation one gets $2ab+2bc+2ac=0$.
If the triangle is isosceles but not equilateral (e.g. $a=b\ne c$), the second and third equation coalesce into $(a+c)^2=2a^2$, which gives a solution: $$ {a\over c}=1+\sqrt2. $$ Finally, we also have the trivial solution $a=b=c$.