# How to find Chord BD in diagram below?

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?

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 $$BD$$.

• Thanks- How are triangles BCD and AOE similar? which rule? Dec 11, 2020 at 11:41
• There is a typo. It is $BOE$ not $AOE$. They are right angle triangles, and the angle at $B$ is the same Dec 11, 2020 at 11:44
• @Andrei - Super. thats exactly threw me off- AOE should be BOE. Dec 11, 2020 at 11:45
• Sorry for the typo... it is difficoult without the name of the points in the picture. Dec 11, 2020 at 11:47
• @TitoEliatron Actually we can use your logic of similarity of BOE and BCD to get BC/BO = CD/OE to get CD as (24/18)*6 = 8 . Then BD = Sqrt(576 - 64) = Sqrt(512) =approx 22.6274 units Dec 11, 2020 at 11:54