label intersection of BC and circle E connect DE and AE. See the solution?
AB=AD+DB=2AD
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.)
so AB=2AC=2x6=12
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!
The fact about the right angle and the diameter is correct (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