ICSE Class 10CISCE
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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

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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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