A bicentric quadrilateral ABCD is inscribed in the circle C(O;R) and circumscribes the circle C(I;r). Let E be the intersection point of the lines AB and CD and let F be the intersection point of the lines BC and AD. Prove that OI is perpendicular on EF.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

    

 

 Demonstration:

 

      Let V be the intersection point of the diagonals AC and BD. We show that: i) , ii) , iii) V, I, O are collinear points.  i), ii) and  iii) .

 

i) Let’s prove that . We prove that V is the orthocenter of the .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let’s prove that . Let S be the intersection point of the lines FV and EO, M - midpoint of the side AB and N - midpoint of the side CD.

We construct ,  and .

We construct  and  

Let’s prove that

The last relation is true.

Thus  and similarly

 

 

ii) Let’s prove that .

   

 

 

 

 

 

 

 

 

 

 

Let M, N, P, Q be the  tangent points of the sides of ABCD to the circle C(I;r).

According to Newton’s theorem it follows that MN, PQ, AC and EF are concurent in point S,

and NP, QM, BD and EF are concurent in point T. It follows that  S, T, E, F are

collinear points.               (1)

 

According to Newton’s theorem it follows that MP, NQ, AC and BD are concurrent in V

. Since MNPQ is cyclic quadrilateral, ,

,  and what shown in i)

(1) and (2)

 

iii) Let’s prove that V, I and O are collinear points.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let’s consider ,,,

, M-midpoint of the AC and  N-midpoint of the BD.

                 

Analogous

Since BDKL-cyclic quadrilateral

It is easy to notice that OMVN-cyclic quadrilateral

Since M-midpoint of the AC and N-midpoint of the BD, according to Newton’s theorem

(The center of the circle inscribed into a quadrilateral lies on the line joining the midpoints of

the latter’s diagonals)  

(1), (2) and (3)