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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?

enter image description here

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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 $BD$.

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  • $\begingroup$ Thanks- How are triangles BCD and AOE similar? which rule? $\endgroup$
    – goldenmean
    Dec 11, 2020 at 11:41
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    $\begingroup$ There is a typo. It is $BOE$ not $AOE$. They are right angle triangles, and the angle at $B$ is the same $\endgroup$
    – Andrei
    Dec 11, 2020 at 11:44
  • $\begingroup$ @Andrei - Super. thats exactly threw me off- AOE should be BOE. $\endgroup$
    – goldenmean
    Dec 11, 2020 at 11:45
  • $\begingroup$ Sorry for the typo... it is difficoult without the name of the points in the picture. $\endgroup$ Dec 11, 2020 at 11:47
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    $\begingroup$ @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 $\endgroup$
    – goldenmean
    Dec 11, 2020 at 11:54

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