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  1. $ABCDEF$ is a circumscribed hexagon. It is known that $AB=2 BC=3 CD=2 DE=5 EF=7.$ Find $FA$.

I can't find any theorem relating to polygons only for quadrilaterals that is the sum of opposite sides must be equal to the sum of the remaining sides. For polygons what's opposite of what?

  1. $ABCDE$ is a regular pentagon.

    What is the angle between $AC$ and $BE$?

What is the angle between $AC$ and the tangent line to the circumcircle of this pentagon at $D$

I can't visualize the 2nd problem at all. When I draw $AC$ and $BE$ I have 2 angles and I don't know which one they refer to. The 2nd part of the problem I cant see whats going on.

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  • $\begingroup$ It's doesnt matter which angle between $AC$ and $BE$ you find - $\alpha$ or $2\pi - \alpha$ but usually the smaller one is considered as "angle between lines". $\endgroup$
    – Andrei Rykhalski
    Oct 13, 2014 at 6:50
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    $\begingroup$ Your claim about quadrilaterals that the "sum of opposite sides must be equal to the sum of the remaining sides" is not true. $\endgroup$
    – Mike Pierce
    Oct 13, 2014 at 6:57
  • $\begingroup$ Are you sure that's what I'm reading in the book. It says AB+CD=BC+DA and it says what I wrote... $\endgroup$
    – adam
    Oct 13, 2014 at 7:04

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  1. Hint: enter image description here $AB+CD+EF=BC+DE+FA$
  2. The sum of the interior angles of a polygon is given by the formula $180(n-3)$ (n is the number of sides).Since $ABCDE$ is a regular pentagon $\angle A= \angle B=\angle C=\angle D= \angle E=\frac{180 (5-2)}{5}=108$ Since $EAB$ and $ABC$ are isosceles triangles then $\angle BEA= \angle CEB =\angle CAB =\frac{180-108}{2}=36.$

Let $O$ be point of intersection of lines $AC$ and $BE.$ In the isosceles triangle $AOB$ we can find $ \angle AOB=108$

I am sorry for my bad English.

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