# Trouble in this circle question

Question says:

In the given figure diameter of the circle is $3$ cm. AB & MN are two diameter such that MN is perpendicular to AB. In addition CG is parallel to MN. $AE : EB = 1 : 2$ and DF is perpendicular to MN and $NL : ML = 1 : 2$, find the length of DH.

Using the ratios, I could formulate that $AE = 1 cm$ and $EB = 2cm$ and $NL = 1cm$ and $ML = 2cm$.
I could also find that $MO = A0 = BO = ON = 1.5cm$ Also, EOLH becomes a square of side 0.5 cm.

But can't figure out how to find DH.

Thanks a lot.

• Are you willing to use Cartesian coordinates and algebra, or are you looking for a purely geometric argument? – Andrew D. Hwang Jan 28 '14 at 12:57
• I think you meant "$\;CG\;$ is parallel to $\;MN\;$" ... – DonAntonio Jan 28 '14 at 13:13
• @user86418 Any. But it would be interesting to see how this problem is solved in both the methods. – Gaurang Tandon Jan 28 '14 at 13:14
• @DonAntonio Thanks! Corrected. – Gaurang Tandon Jan 28 '14 at 13:14

$OD$ is a radius of the circle as well, so has length $1.5$. $OL$ has length $0.5$. So by Pythagoras $DL$ has length $\sqrt{1.5^2 - 0.5^2} = \sqrt{2}$. And $HL$ has length $0.5$ so $DH$ has length $\sqrt 2 - 0.5$.

• Very nice approach! – Gaurang Tandon Jan 28 '14 at 13:44

Triangle $MDN$ is right-angled, so triangles $DLN$ and $MLD$ are similar. So $ML/LD = LD/LN$. This gives you $LD$, and so $DH$.

First of all we want to find the angle $DOL$. Using cosine of that angle,

$cos(DOL) = \frac{adjacent}{hypotenuse} = \frac{OL}{OD} = \frac{0.5}{1.5}$

Therefore $DOL = sin^{-1}(\frac{0.5}{1.5}) = 0.33983690945412193$.

Now we know that

$sin(DOL) = \frac{opposite}{hypotenuse} = \frac{DL}{OD} = \frac{DL}{1.5}$

Rearranging,

$DL = sin^{-1}(DOL) * 1.5 = 1.4142135623730949$

So $DH = DL - HL = 1.4142135623730949 - 0.5 = 0.91421356237309492$

Sketch of coordinate approach: The circle of radius $3r$ centered at the origin has equation $x^2 + y^2 = 9r^2$, and the lines $CG$ and $DF$ have respective equations $x = -r$ and $y = -r$. The coordinates of $D$ and $H$ are easy to express in terms of $r$.