+1725 Let $AC$ be a diameter of a circle $\omega$ of radius $1$, and let $D$ be the point on $AC$ such that $CD = \frac{1}{5}$. Let $B$ be the point on $\omega$ such that $DB$ is perpendicular to $AC$, and let $E$ be the midpoint of $DB$. Compute length $AE$.
+129771 
AO = 1 = CO
AD = AO + CO - CD = 1 + 1 - 1/5 = 2 -1/5 = 9/5
DC = 1/5
AC perpendicular to BF ....so chord BF is bisected by AC
So.....BD = DF
Intersecting Chord Theorem
AD * DC = BD * DF
AD * DC = BD * BD
AD * DC = BD^2
(9/5) (1/5) = BD^2
9/25 = BD^2
BD = 3/5
DE = (1/2)(BD) = (1/2)(3/5) = 3/10
Triangle AED is right
Pythagorean Theorem
AE^2 = AD^2 + DE^2
AE^2 = (9/5)^2 + (3/10)^2
AE^2 = 81/25 + 9/100
AE^2 = [324 + 9 ] / 100
AE^2 = 333/100
AE = sqrt (333) / 10 ≈ 1.82
