Find the length of the chord BD

Question

Two chords AB and AC are drown in a circle with diameter AD. Find the length of the chord BD when angle BAC = 60°, BE is perpindicular to AC, and EC = 3 cm

NOTE: answers can include surds

Apart from being not drawn to scale, the real figure will be much different, it seems

I have also calculated the length of AE to be sqrt(3), however I can not guarantee that it is correct

Answer 1

Hints:
1. $DC \parallel BE$.
2. Angle between $BD$ and $AC$ is $30^\circ$.
3. Translate BD by DC which moves B to X and D to C. Then, $XC=DB$ and $XCE$ is a $30^\circ-60^\circ-90^\circ$ triangle with a base length of $3$.

Answer 2

We'll extend BE to touch the circle in point X. Then we'll draw BD. Note that angle ABE is 30 degrees. ABD is drawn in a semi-circle so ABD.is 90 degrees. So EBD is 60 degrees The arc BDC is 2^BAC which is 120 degrees. And the arc DCX is 2^DBX which is 120 degrees. So the arc BDC equals the arc DCX And by removing the arc DC from both sides: The arc BD equals the arc CX So its cord BD equals the other's cord CX. So we wanna find CX: Note that AEB is a 30-60 right triangle So AE is G and BE is G multiplied by the square root of three. AC AND BX ARE INTERSECTING CORDS IN A CIRCLE SO BE ^Ex =AE ^EC SO G MULTIPIED BY THE SQUARE ROOT OF THREE ^EX = 3G The G goes with the G So XE is three over the square root of three CEX IS A RIGHT TRIANGLE SO XC =BD = The square root of twelve.