The bases BC and AD of the trapezoid ABCD are 5 and 20, respectively, BD = 10. Prove that triangles CBD and ADB are similar

ABCD is a trapezoid, therefore, AD || BC.
∠CBD = ∠ADB (as these are cross-angles for parallel lines AD and BC).
Consider the relationship of the parties:
BC / BD = 5/10 = 1/2
BD / AD = 10/20 = 1/2
Then, according to the second similarity sign of triangles, triangles CBD and ADB are similar

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