By similar triangles, one can prove that the common tangent will intersect with the x-axis at[-a(b+a)/(b-a),0] Further the length of the common tangent will be 2√(ab) which gives us 2C = ∏ab while A=∏a^2/2 & B =∏b^2/2 and thus A + B + 2C = ∏a^2/2 + ∏b^2/2 + ∏ab = (∏/2)(a+b)^2 -----(1) whereas D=∏((a+b)/2)^2/2 or 4D= (∏/2)(a+b)^2=A+B+2C by equation (1). Ajit
By similar triangles, one can prove that the common tangent will intersect with the x-axis at[-a(b+a)/(b-a),0] Further the length of the common tangent will be 2√(ab) which gives us 2C = ∏ab while A=∏a^2/2 & B =∏b^2/2 and thus A + B + 2C = ∏a^2/2 + ∏b^2/2 + ∏ab = (∏/2)(a+b)^2 -----(1) whereas D=∏((a+b)/2)^2/2 or 4D= (∏/2)(a+b)^2=A+B+2C by equation (1).
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