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ILL 1969 Problem 10

    (BUL4)(BUL 4) Let MM be the point inside the right-angled triangle ABC(∠C=90∘)ABC (\angle C = 90^{\circ}) such that ∠MAB=∠MBC=∠MCA=ϕ.\angle MAB = \angle MBC = \angle MCA =\phi. Let Ψ\Psi be the acute angle between the medians of ACAC and BC.BC. Prove that sin⁡(ϕ+Ψ)sin⁡(ϕ−Ψ)=5.\frac{\sin(\phi+\Psi)}{\sin(\phi-\Psi)}= 5.

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