# Let triangle $ABC$ have $AB = AC$… Prove that $KB = KD$

Let triangle $ABC$ have $AB = AC$. In a straight line perpendicular to $AC$ at $C$ take $D$ so that two points $B$ and $D$ are different from the direction of $AC$. Let $K$ be the intersection of the straight line through $B$ perpendicular to $AB$ and the line through the midpoint $M$ of the $CD$ and perpendicular to $AD$. Prove that $KB = KD$

A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$
Since $AD\bot MK$ we have