Q. 105.0( 1 Vote )

# 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.

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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