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IMO Team Selection Test 3

Geometry Difficulty 8.1 Shortlist Prove it Netherlands

Let ABCDABCD be a parallelogram and let MM be the intersection of its diagonals. The circumcircle of △ABM\triangle ABM intersects the line segment ADAD in E≠AE \ne A and the circumcircle of △EMD\triangle EMD intersects the line segment BEBE in the point F≠EF \ne E.
Prove that ∠ACB=∠DCF\angle ACB = \angle DCF.

Solution

We first show that CBFDCBFD is a cyclic quadrilateral. Note that
∠BCD=∠BAD(parallelogram)=∠BAE=180∘−∠EMB(EABM is cyclic)=∠EMD=∠EFD(inscribed angle theorem in EFMD)=180∘−∠BFD \angle BCD = \angle BAD \quad \text{(parallelogram)} = \angle BAE = 180^\circ - \angle EMB \quad \text{(EABM is cyclic)} = \angle EMD = \angle EFD \quad \text{(inscribed angle theorem in EFMD)} = 180^\circ - \angle BFD
Therefore CBFDCBFD is a cyclic quadrilateral. Using the inscribed angle theorem with respect to cyclic quadrilaterals EABMEABM, EFMDEFMD, and CBFDCBFD when necessary, we see that
∠ACD=∠CAB=∠MAB=∠MEB=∠MEF=∠MDF=∠BDF=∠BCF. \angle ACD = \angle CAB = \angle MAB = \angle MEB = \angle MEF \\ = \angle MDF = \angle BDF = \angle BCF.
So ∠ACF+∠FCD=∠ACD=∠BCF=∠BCA+∠ACF\angle ACF + \angle FCD = \angle ACD = \angle BCF = \angle BCA + \angle ACF. If we subtract ∠ACF\angle ACF from this we find that ∠ACB=∠DCF\angle ACB = \angle DCF, as required. □\square

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