# Intersecting diameter and chord

Mathematics Asked on January 1, 2022

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 $$CE^2=AEcdot EB.$$ Also, $$AMcdot MB=CMcdot MD$$ and we obtain the following system: $$(2x)^2-(2y)^2=1cdot(2y+7y+49)$$ and $$(1+2y)(7y+49)=2xcdot7x.$$ 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 $$CD=39$$.

Answered by Michael Rozenberg on January 1, 2022