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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 $KB = KD$

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We will use this property:

A line $XY$ is perpendicular to a line $AB$ iff $$AX^2-AY^2 =BX^2-BY^2$$

Since $AD\bot MK$ we have

\begin{eqnarray} KD^2 &=& KA^2 +MD^2-MA^2\\ &=& \underbrace{AB^2+BK^2}_{=KA^2} +MD^2- MA^2\\ &=& {AB^2}+BK^2 +MD^2- (\underbrace{AC^2+MC^2}_{=MA^2})\\ &=& BK^2 \end{eqnarray}

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