label intersection of BC and circle E connect DE and AE. See the solution?
Meanwhile AD=AC=6 (the base of right angled triangle in a circle is on the diameter, congruent triangles. draw that line and you'll see what I mean.)
I cannot be bothered to write a full proof right now as this question is quite simple.'
Oh! This is interesting, it's a 30 60 90 triangle. And the triangles inside the circle are also 30 60 90. Beautiful!
Thales' theorem), but it does not follow that triangles ADE and ACE are congruent and that AD = AC.
Instead, ED is both a median and an altitude in triangle ABE, so triangle ABE is...
BEC perpendicular to AC given AC is adiameter (subtends arc of 180
Triangle AEC 10^2 -6^2= EC^2 EC =8
ED is perpendicular bisector of AB (AD=DB given) and angle ADE =90 (subtends arc of dia)
AE=EB=10 BC = 18 18 ^2 +6^2 = AB^2 AB=18.97