# Cyclic Pentagon 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 $BE$?

• "distance from point $A$ to $BC$ is $a$". This is the perpendicular distance or what? – EuYu Nov 7 '12 at 22:34
• @amWhy the distance is the perpendicular distance, yes, in all cases. – George Krasilnikov Nov 7 '12 at 23:16
• @EuYu the distance is the perpendicular distance, yes, in all cases – George Krasilnikov Nov 7 '12 at 23:17
• @AmWhy not necessarily. Nothing in the question suggests that they are. – George Krasilnikov Nov 7 '12 at 23:46
• 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. – 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 $$d(A,BC)d(A,DE)=4r^2|\sin(\beta/2)\sin(\gamma/2)\sin(\delta/2)\sin(\epsilon/2)|=d(A,BE)d(A,CD)$$
• If you know $xy=zw$ and you know $x,y$ and $z$, how do you find $w$? – Robert Israel Nov 8 '12 at 0:11