#### 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`

Is there an error in this question or solution?

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.