A Circle is drawn with center as A, radius as 12 units. A smaller incircle is drawn as shown with center E and which passes through A. BD is the tangent to the smaller circle. Find BD?
EDIT: Line DC need not be tangent to smaller circle.
I tried following thigs:
1] Constructed CD , making angle BDC as 90 = > Angle subtended by semicircle.
2] Using Pythagoras I see $BD^2 + CD^2 = 24^2$
3] I tried many things, but cannot find CD ?
e.g. I joined points A and D in hope of getting angle DAC which might help in finding Chord CD , which in turn could help in getting BD, but I could not progress in that direction?
How could I find length of BD ? IS there any other approach to solve this problem than what I have tried so far?
Best Answer
If you call $E$ te tangent point of $BD$ and the small circle, AND if you call $O$ the center of the small circle, you have that the triangles $BOE$ and $BCD$ are similar (recall that $BD\perp OE$ since $OE$ is the radius at the tangent point and $BD$ is the tangent line).
So $\frac{BE}{BO}=\frac{BD}{BC}$.
BUT, $BO=12+6=18.$ $BC=24$ and $BE^2+OE^2=BO^2$, whence $BE=\sqrt{BO^2-OE^2}=\sqrt{18^2-6^2}=\sqrt{288}=12\sqrt2$.
Now, it is easy to determine the length of $BD$.