olymmath

OMK 2019 B2

    Given a parallelogram ABCDABCD , a point M is chosen such that ∠DAC=∠MAC\angle DAC=\angle MAC and ∠CAB=∠MAB.\angle CAB=\angle MAB. Prove AMBM=(ACBD)2\frac{AM}{BM}=\left(\frac{AC}{BD}\right)^2

    Dataset text AI-MO AoPS. Quoted with source.

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