AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.
Is it even possible to prove it? I mean I think BC is less that half of BD...
angle ACB = 90 degrees....because it is subtended by a diameter, by a semi-circle.
So, angle ABC = 180 -30 -90 = 60 degrees.
angle ABD is a straight angle, so it is 180 degrees.
And so, angle CBD = 180 -angle ABC = 180 -60 = 120 degrees.
Minor arc BC = 2(angle BAC) = 2(30) = 60 degrees....an inscribed angle is half of the intercepted arc in measure.
So, angle BCD = (1/2)(60) = 30 degrees....a tangent-chord angle is half of the intercepted arc in measure.
So, angle BDC = 180 -120 -30 = 30 degrees.
Hence, triangle CBD is an isosceles triangle, whose equal sides are BC and BD.
Therefore, BC = BD.