A diameter $AB$ and a chord $CD$ of a circle $k$ intersect at $M.$ $CE$ and $DF$ are perpendiculars from $C$ and $D$ to $AB$. $(A,E,M,F,B$ lie on AB in that order$)$. What is the length of $CD$ if $AE=1,FB=49$ and $MC:MD=2:7$?
How do I approach the given problem? I would be very grateful if you could give me some hints and tips to follow. I see that the triangles $CEM$ and $DFM$ are similar and $\dfrac{MC}{MD}=\dfrac{CE}{DF}=\dfrac{EM}{FM}=\dfrac{2}{7}.$
Let $CM=2x$ and $EM=2y$.
Thus, $$MF=7y,$$ $$MD=7x$$ and since $\measuredangle ACB=90^{\circ}$, we obtain $$CE^2=AE\cdot EB.$$ Also, $$AM\cdot MB=CM\cdot MD$$ and we obtain the following system: $$(2x)^2-(2y)^2=1\cdot(2y+7y+49)$$ and $$(1+2y)(7y+49)=2x\cdot7x.$$ The last equality it's $$(1+2y)(y+7)=2x^2$$ and we can substitute $2x^2$ in the first equation.
Thus $$2(1+2y)(y+7)-4y^2=9y+49.$$ Can you end it now?
I got $CD=39$.