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Let ABC be an isosceles triangle AB=ACAB=ACAB=AC with ∠BAC=20∘\angle BAC = 20^{\circ} ∠BAC=20∘. Point DDD is on side ACACAC such that ∠DBC=60∘\angle DBC = 60^{\circ} ∠DBC=60∘. Point EEE is on side ABABAB such that ∠ECB=50∘\angle ECB = 50^{\circ} ∠ECB=50∘. Find the measure of ∠EDB\angle EDB∠EDB in degrees.
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