93. Given an isosceles triangle \(ABC, |AB| = |BC|, AD\) is the angle bisector. A perpendicular line to \(AD\) at point \(D\) intersects the extension of \(AC\) at point \(E\); the feet of the perpendiculars dropped from \(B\) and \(D\) to \(AC\) are \(M\) and \(N\). Find \(|MN|\), if \(|AE| = a\).
93. Let $P$ be the intersection point of line $DE$ with $AB$, and $K$ be a point on $AB$ such that $KD \| AC$. $\triangle AKD$ is isosceles with $\widehat{(KDA} = \widehat{DAC} = \widehat{DAK})$. Therefore, $KD$ is the median in a right triangle, and $|MN| = \frac{1}{2}|KD| = \frac{1}{4}|AP| = \frac{1}{4}|AE| = \frac{1}{4}a$.