In a circle, with centre O ,a diameter AB and a chord AD are drawn . Another circle is drawn with AO as diameter to cut AD at C. Prove that : BD = ` 2 xx OC `

Answer

To prove that \( BD = 2 \times OC \) in the given configuration, we can follow these steps:

### Step 1: Identify the triangles
We have two triangles: triangle \( OAC \) and triangle \( BAD \).

### Step 2: Establish right angles
Since \( AB \) is a diameter of the circle, angle \( BDA \) is \( 90^\circ \) (angle in a semicircle). Similarly, since \( AO \) is also a diameter of the smaller circle, angle \( OCA \) is also \( 90^\circ \).

### Step 3: Identify common angles
In both triangles, angle \( OAD \) is common. Therefore, we have:
- \( \angle BDA = 90^\circ \)
- \( \angle OCA = 90^\circ \)
- \( \angle OAD = \angle BAD \)

### Step 4: Use Angle-Angle (AA) similarity
Since both triangles share a common angle and each has a right angle, by the Angle-Angle (AA) similarity criterion, we can conclude that:
\[ \triangle OAC \sim \triangle BAD \]

### Step 5: Set up the proportion of sides
From the similarity of triangles, we can write the following proportions:
\[ \frac{BA}{OA} = \frac{BD}{OC} \]
\[ \frac{AD}{AC} \]

### Step 6: Substitute the lengths
Let \( R \) be the radius of the larger circle. Then:
- \( BA = 2R \) (diameter of the larger circle)
- \( OA = R \) (radius of the smaller circle)

Substituting these values into the proportion gives:
\[ \frac{2R}{R} = \frac{BD}{OC} \]

### Step 7: Simplify the equation
This simplifies to:
\[ 2 = \frac{BD}{OC} \]

### Step 8: Rearrange to find \( BD \)
Multiplying both sides by \( OC \) gives:
\[ BD = 2 \times OC \]

### Conclusion
Thus, we have proved that \( BD = 2 \times OC \).

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