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AD is bisector of ∠ A it divided the BC in ratio of AB : AC = 10 : 15 = 2 : 3 and by PGT = AC = 5√13 DC = 3√13 Let [C = Q0] Now apply sinθ in ∆Abc and in ∆ EDC we get = [sinθ = 10/5√13 = DE/3√13 DE = 6cm
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