Question 326.
Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in C. If now DE be drawn perpendicular to AB, and if AC, DE meet in F, then what is the ratio of DF and EF ?
Answer 326.
Let the equation of the circle be x^2 + y^2 = r^2
Let D = (rcosθ, rsinθ)
Eqn. of tangent at B is x = r ... (1)
Eqn. of tangent at D is xcosθ + ysinθ = r ... (2)
Solving eqns. (1) and (2),
C = [r, r(1-cosθ)/sinθ] = [r, rtan(θ/2)]
Δs ABC and AEF are similar
=> FE
= BC * (AE/AB)
= rtan(θ/2) * (r + rcosθ)/2r
=(1/2) r tan(θ/2) (1 + cosθ)
= (1/2) r tan(θ/2) * 2cos^2 (θ/2)
= (1/2) r sinθ ... (3)
DF
= DE - FE
= rsinθ - (1/2) rsinθ
= (1/2) rsinθ ... (4)
From eqns. (3) and (4),
DF = FE
=> DE : EF = 1.
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Let AB be the diameter of a semicircle, and let the tangents to it at B and at any other point D on it meet in C. If now DE be drawn perpendicular to AB, and if AC, DE meet in F, then what is the ratio of DF and EF ?
Answer 326.
Let the equation of the circle be x^2 + y^2 = r^2
Let D = (rcosθ, rsinθ)
Eqn. of tangent at B is x = r ... (1)
Eqn. of tangent at D is xcosθ + ysinθ = r ... (2)
Solving eqns. (1) and (2),
C = [r, r(1-cosθ)/sinθ] = [r, rtan(θ/2)]
Δs ABC and AEF are similar
=> FE
= BC * (AE/AB)
= rtan(θ/2) * (r + rcosθ)/2r
=(1/2) r tan(θ/2) (1 + cosθ)
= (1/2) r tan(θ/2) * 2cos^2 (θ/2)
= (1/2) r sinθ ... (3)
DF
= DE - FE
= rsinθ - (1/2) rsinθ
= (1/2) rsinθ ... (4)
From eqns. (3) and (4),
DF = FE
=> DE : EF = 1.
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