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Geometry Difficulty 6.7 National Olympiad Prove it Bulgaria

Problem:

The diagonals ACAC and BDBD of a cyclic quadrilateral ABCDABCD with circumcenter II intersect at a point EE. If the midpoints of segments ADAD, BCBC and IEIE are collinear, prove that AB=CDAB = CD.

Solution

Solution:

Denote the midpoints of ADAD and BCBC by MM and NN, respectively. Without loss of generality we may assume that II lies in the interior of AMNBAMNB and EE lies in the interior of MDCNMDCN. Then we have
SMIN=SAMNB−SABI−SAMI−SBNI=SAMNB−SABI−12(SADI+SBCI)=SAMNB−12SABI−12(SABCD−SCDI) \begin{aligned} S_{MIN} & = S_{AMNB} - S_{ABI} - S_{AMI} - S_{BNI} \\ & = S_{AMNB} - S_{ABI} - \frac{1}{2}\left(S_{ADI} + S_{BCI}\right) \\ & = S_{AMNB} - \frac{1}{2} S_{ABI} - \frac{1}{2}\left(S_{ABCD} - S_{CDI}\right) \end{aligned}

Figure 1

Analogously, SMEN=SMDCN−12SDCE−12(SABCD−SABE)S_{MEN} = S_{MDCN} - \frac{1}{2} S_{DCE} - \frac{1}{2}\left(S_{ABCD} - S_{ABE}\right). Now using SMIN=SMENS_{MIN} = S_{MEN} we get

SAMNB−12SABI−12(SABCD−SCDI)=SMDCN−12SDCE−12(SABCD−SABE)S_{AMNB} - \frac{1}{2} S_{ABI} - \frac{1}{2}\left(S_{ABCD} - S_{CDI}\right) = S_{MDCN} - \frac{1}{2} S_{DCE} - \frac{1}{2}\left(S_{ABCD} - S_{ABE}\right),

12SADN+12SABC−12SABI−12(SABCD+12SCDI)=12SADN+12SCBD−12SDCE−12(SABCD+12SABE)SABC−SCBD+SDCE−SABE=SABI−SCDISABE−SCDE+SDCE−SABE=SABI−SCDI \begin{aligned} & \frac{1}{2} S_{ADN} + \frac{1}{2} S_{ABC} - \frac{1}{2} S_{ABI} - \frac{1}{2}\left(S_{ABCD} + \frac{1}{2} S_{CDI}\right) \\ = & \frac{1}{2} S_{ADN} + \frac{1}{2} S_{CBD} - \frac{1}{2} S_{DCE} - \frac{1}{2}\left(S_{ABCD} + \frac{1}{2} S_{ABE}\right) \\ & S_{ABC} - S_{CBD} + S_{DCE} - S_{ABE} = S_{ABI} - S_{CDI} \\ & S_{ABE} - S_{CDE} + S_{DCE} - S_{ABE} = S_{ABI} - S_{CDI} \end{aligned}

Therefore SABI=SCDIS_{ABI} = S_{CDI}, whence AB=CDAB = CD.

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