If the chord ab of a circle is parallel to the tangent at c, then prove that ac=bc
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Step-by-step explanation:
Let DAE be tangent at A to the circumcircle of ΔABC. In ΔABC, AB = AC (Given) ∴ ∠ACB = ∠ABC ----- (1) (Angles opposite to equal sides are equal) According to alternate segment theorem, the angle between the tangent and chord at the point of contact is equal to the angles made by the chord in the corresponding alternative segment. DAE is the tangent and AB is the chord. ∴ ∠DAB = ∠ACB ------ (2) From (1) and (2), we have ∠ABC = ∠DAB ∴ DE is parallel to BC (If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel) Therefore, the tangent at A to the circumcircle of ΔABC is parallel BC.
anildeshmukh:
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