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IMO Hk TST

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Let MM be the intersection point of the diagonals ACAC and BDBD of a convex quadrilateral ABCDABCD. The bisector of ∠ACD\angle ACD meets BABA extended at KK. If MA⋅MC+MA⋅CD=MB⋅MDMA \cdot MC + MA \cdot CD = MB \cdot MD, show that ∠BKC=∠CDB\angle BKC = \angle CDB.

Solution

Construct a point EE on the extension of ACAC such that CD=CECD = CE. Using the condition, we find that
MB⋅MD=MA⋅MC+MA⋅CD=MA(MC+CE)=MA⋅ME. \begin{aligned} MB \cdot MD &= MA \cdot MC + MA \cdot CD \\ &= MA(MC + CE) \\ &= MA \cdot ME. \end{aligned}
This implies A,B,E,DA, B, E, D are concyclic.

Now, since ∠CDE=∠CED=12∠DCA=∠KCA\angle CDE = \angle CED = \frac{1}{2} \angle DCA = \angle KCA, we obtain
∠BKC=∠BAC−∠KCA=∠BDE−∠CDE=∠BDC \angle BKC = \angle BAC - \angle KCA = \angle BDE - \angle CDE = \angle BDC
as desired.

Figure 1

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