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 x AD = AB²

(ii) BD² = AD x DC.

(i) In ∆ABC, we have

∠B = 90° and BC is the diameter of the circle.

Hence, AB is the tangent to the circle at B.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AB² = AD x AC.

Hence, proved that AB² = AD x AC.

(ii) From figure,

∠BDC = 90° [Angle in a semi-circle is a right angle.]

From figure,

⇒ ∠ADB + ∠BDC = 180° [Linear pairs]

⇒ ∠ADB + 90° = 180°

⇒ ∠ADB = 180° - 90°

⇒ ∠ADB = 90°

In ∆ADB,

⇒ ∠ADB + ∠A + ∠ABD = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ABD = 180°

⇒ ∠A + ∠ABD = 90° ……………(1)

In ∆ABC, ∠ABC = 90°.

⇒ ∠ABC + ∠A + ∠ACB = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ACB = 180°

⇒ ∠A + ∠ACB = 90° ……………(2)

From (1) and (2),

⇒ ∠A + ∠ABD = ∠A + ∠ACB

⇒ ∠ABD = ∠ACB.

From figure,

⇒ ∠ACB = ∠BCD

∴ ∠ABD = ∠BCD

Now in ∆ABD and ∆CBD, we have

∠BDA = ∠BDC [Both equal to 90°]

∠ABD = ∠BCD

Hence, ∆ABD ~ ∆CBD by AA postulate.

We know that,

Ratio of corresponding sides of similar triangles are same.

$\dfrac{BD}{DC} = \dfrac{AD}{BD}$

∴ BD² = AD x DC.

Hence, proved that BD² = AD x DC.