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Consider the above pentagon. Suppose that the distance from point $A$ to $BC$ is $a$, the distance from $A$ to $CD$ is $b$, and the distance from $A$ to $DE$ is $c$. In terms of this, how can we find the distance from $A$ to $BE$?

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    $\begingroup$ "distance from point $A$ to $BC$ is $a$". This is the perpendicular distance or what? $\endgroup$ – EuYu Nov 7 '12 at 22:34
  • $\begingroup$ @amWhy the distance is the perpendicular distance, yes, in all cases. $\endgroup$ – George Krasilnikov Nov 7 '12 at 23:16
  • $\begingroup$ @EuYu the distance is the perpendicular distance, yes, in all cases $\endgroup$ – George Krasilnikov Nov 7 '12 at 23:17
  • $\begingroup$ @AmWhy not necessarily. Nothing in the question suggests that they are. $\endgroup$ – George Krasilnikov Nov 7 '12 at 23:46
  • $\begingroup$ Actually, your particular image suggests they are equivalent; and since you left out the important specification of what you mean by "distance" of A to ______, I thought it best to have you clarify whether or not the lengths BC, CD, DE were equivalent, etc. $\endgroup$ – Namaste Nov 8 '12 at 0:39

Hint: express each distance in terms of the radius of the circle and the cosines of the angles subtended at the centre by $AB$, $AC$, $AD$ and $AE$. If I'm not mistaken, you should find that the product of two of the distances is equal to the product of the other two.

EDIT: in fact, if $\beta$ and $\gamma$ are the angles subtended at the centre by $AB$ and $AC$ and the radius is $r$, I find that the distance from $A$ to $BC$ is $2 r |\sin(\beta/2) \sin(\gamma/2)|$. Similarly of course for the other distances.

EDIT (incorporating comment as requested): Given that $d(A,BC)=2r|\sin(\beta/2)\sin(\gamma/2)|$ and similarly $d(A,CD)=2r|\sin(\gamma/2)\sin(\delta/2)|$, $d(A,DE)=2r|\sin(\delta/2)\sin(\epsilon/2)|$ and $d(A,BE)=2r|\sin(\beta/2)\sin(\epsilon/2)|$, we have $$d(A,BC)d(A,DE)=4r^2|\sin(\beta/2)\sin(\gamma/2)\sin(\delta/2)\sin(\epsilon/2)|=d(A,BE)d(A,CD)$$

  • $\begingroup$ That's what I'm getting, but my derivation a bit cluttered. Considering how nice the relation is, I keep thinking there should be an elegant path to it. $\endgroup$ – Blue Nov 7 '12 at 23:53
  • $\begingroup$ @Robert How will that help me to find the distance between A and BE? $\endgroup$ – George Krasilnikov Nov 8 '12 at 0:07
  • $\begingroup$ If you know $xy=zw$ and you know $x,y$ and $z$, how do you find $w$? $\endgroup$ – Robert Israel Nov 8 '12 at 0:11
  • $\begingroup$ @Robert but BE is not any of AB, AC, AD, AE. $\endgroup$ – George Krasilnikov Nov 8 '12 at 0:12
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    $\begingroup$ @If you can add your comments into a full solution as part of your answer, that will be great- then I can mark it as correct. $\endgroup$ – George Krasilnikov Nov 8 '12 at 13:37

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