In the given figure, ∠ABC = 90° and BC is diameter of the given circle. Show that :

(i) AC × AD = AB²

(ii) AC × CD = BC²

(i) As, BC is the diameter.

We know that,

Angle in a semicircle is a right angle.

∴ ∠BDC = 90°

From figure,

⇒ ∠BDC + ∠BDA = 180° [Linear Pair]

⇒ 90° + ∠BDA = 180°

⇒ ∠BDA = 180° - 90°

⇒ ∠BDA = 90°.

As, AB is the tangent and BC is diameter and tangent at any point and line from that point to center are perpendicular to each other.

⇒ ∠ABC = 90°

In △ABC and △ABD,

∠ABC = ∠ADB (Both equal to 90°)

∠BAD = ∠BAC (Common)

∴ △ABC ~ △ABD

In similar triangles,

Ratio of corresponding sides are in equal proportion.

$\therefore \dfrac{AC}{AB} = \dfrac{AB}{AD} \\[1em] \Rightarrow AB^2 = AC \times AD.$

Hence, proved that AB² = AC × AD.

(ii) In △ABC and △BDC,

⇒ ∠ABC = ∠BDC (Both equal to 90°)

⇒ ∠BCA = ∠BCD (Common)

∴ △ABC ~ △BDC

In similar triangles,

Ratio of corresponding sides are in equal proportion.

$\therefore \dfrac{AD}{BC} = \dfrac{BC}{CD} \\[1em] \Rightarrow BC^2 = AD \times CD.$

Hence, proved that BC² = AD × CD.