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Abc is a Right Triangle with Angle B = 90°, a Circle with Bc as Diameter Meets Hypotenuse Ac at Point D. Prove That: (I) Ac × Ad = Ab^2 (Ii) Bd^2 = Ad × Dc - ICSE Class 10 - Mathematics

ConceptTangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments

Question

ABC is a right triangle with angle B = 90°, A circle with BC as diameter meets hypotenuse AC at point D. prove that:
(i) AC × AD = AB^2
(ii) BD^2 = AD × DC

Solution

i) In Δ ABC,
∠B = 90° and BC is the diameter of the circle.
Therefore, AB is the tangent to the circle at B.
Now, AB is tangent and ADC is the secant
∴ AB^2 = AD × AC

ii) In Δ ADB,
∠D = 90°
∴  ∠A + ∠ABD = 90° ……..(i)
But in ΔABC, ∠B = 90
∴ ∠A + ∠C = 90 …………(ii)
From (i) and (ii)
∠C  =  ∠ABD
Now in ΔABDand ΔCBD

∠BDA = ∠BDA = 90°

∠ABD = ∠BCD

∴ Δ ABD ∼ Δ CBD (AA postulate)

∴"BD"/"DC" = "AD"/"DP"

⇒ BD^2 = AD xx DC

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Solution Abc is a Right Triangle with Angle B = 90°, a Circle with Bc as Diameter Meets Hypotenuse Ac at Point D. Prove That: (I) Ac × Ad = Ab^2 (Ii) Bd^2 = Ad × Dc Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
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