Q10 of 31 Page 218

Two tangents AB and AC drawn from an external Point A of a circle touch the circle at the point B and C. A tangent drawn to a point X lies on minor arc BC intersects AB and AC at the points D and E respectively. Let us prove that perimeter of ΔADE = 2AB.

Capture.JPG


AB and AC are two tangents from external point


Since DE is a variable tangent, it meets the circle at a variable point X


AB = AC (Tangents drawn from an external point to the same circle are always equal)


So we can say


2AB = AB + AC


⇒ 2AB = (AD + DB) + (AE + EC)


⇒ 2AB = (AD + AE) + (DB + EC) …Equation(i)


DB = DX (Tangents drawn from an external point to the same circle are always equal)


XE = EC (Tangents drawn from an external point to the same circle are always equal)


Replacing DB by DX and EC by XE we get


⇒ 2AB = (AD + AE) + (DX + XE)


⇒ 2AB = AD + AE + DE


⇒ 2AB = Perimeter of Δ ADE


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