A circle has 2 equal chords AB&AC.Chord AD bisects BC in E.If AC=12cm& AE=8cm, find AD.?

Answer 1

AB = AC

The point A is midway between B and C on the circle.
AD bisects BC at point E. Therefore AD is a diameter of the circle.

Let O be the center of the circle. AD passes thru point O.

OA is a radius of the circle and so is OC.
Let OA = OC = r.

We have the right triangle AEC with the right angle at E. By the Pythagorean Theorem we have:

(CE)² = (AC)² - (AE)² = 12² - 8² = 144 - 64 = 80

We also have the right triangle CEO with the right angle at E. By the Pythagorean Theorem we have:

(OC)² = (CE)² + (OE)² = 80 + (OE)²
r² = 80 + (OE)²
(OE)² = r² - 80

We also know that:
r = OA = OE + EA = OE + 8
OE = r - 8
(OE)² = (r - 8)² = r² - 16r + 64

Set the two equations equal.
r² - 80 = r² - 16r + 64
16r = 64 + 80 = 144
r = 9

AD is a diameter so
AD = 2r = 2*9 = 18 cm

Answer 2

As far as I can remember "bisect" means to split into two EQUAL parts...If that is correct, then if AE is 8cm then AD would be twice that and thus 16cm...this is all provided bisect means to split equally.