When the problem statement beats most horror stories

[Zawadx]

Let $ABC$ be a triangle such that $AB \neq AC$ and $\angle B \neq 90$. Let $I$ be the incenter of $\triangle ABC$, and $D,E,F$ be the perpendicular feet at $BC, CA, AB$ from $I$ respectively. Let $S$ be the intersection of $AB$ and $DI$, and $T$ be the intersection of $DE$ and the line passing $F$ and perpendicular to $DF$, and $R$ be the intersection of $ST$ and $EF$. Let $P_{ABC}$ be the intersection point of the circle with diameter $IR$ and the incircle of $\triangle ABC$, that is on the other side to $A$ with respect to the line $IR$.

Let $\triangle XYZ$ be an isosceles triangle with $XZ=YZ>XY$ and $W$ be a point on $YZ$ with $WY<XY$. For $K=P_{YXW}$ and $L=P_{ZXW}$ show that $XY \geq 2KL$.

[joydip]

Let $ M$ be the midpoint of $BC$ .$D’$ be the antipode of $ D$ wrt $(I)$.N be the reflection of $D$ wrt $M$.
Then $D’= FT \cap DS$. By pascals theorem on hexagon $FD’D’DEF$ we get $D’R$ is tangent to $(I) $ .$AD’$ is the polar of $ R$ wrt $(I)$.So $AD’ \perp IR $.So $ P\in AD’$.Again $N \in AD’$ (well known). $DP \perp PN$. So $2MP=DN= |AB-AC|$
Let $W_1 \in ZY$ such that $W_1Y=XY$. $J$ be the midpoint of $XW_1$. If $W=W_1$ then as shown above $P_{YXW}=K= J$ .Then $2KL=ZX-ZW_1=ZY-ZW_1= W_1Y=XY$
If $WY < XY$ then $K $ & $ X$ lies on the same side of $YJ$ ( the angle bisector of $\angle XYZ$) .So $\angle LJK > \angle W_1ZY=90^\circ$ .So $2KL > 2JL=XY$.