ABCD is a cyclic quadrilateral in which AB is a diameter , `,BC=CDandangleABD=50^(@)` then `angleDBC` is :

Options

A. `25^(@)`
B. `20^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we will follow these steps:

1. **Identify the Given Information**: We have a cyclic quadrilateral ABCD where AB is a diameter, BC = CD, and ∠ABD = 50°.

2. **Use the Property of Cyclic Quadrilaterals**: Since AB is a diameter, ∠ADB will be a right angle (90°) because the angle subtended by a diameter at the circumference is always 90°.

\[ \angle ADB = 90° \]

3. **Find Angle DAB**: In triangle ADB, we know that the sum of angles in a triangle is 180°. Thus, we can find angle DAB as follows:

\[ \angle DAB + \angle ABD + \angle ADB = 180° \]

Substituting the known values:

\[ \angle DAB + 50° + 90° = 180° \]

Simplifying this gives:

\[ \angle DAB + 140° = 180° \]

\[ \angle DAB = 180° - 140° = 40° \]

4. **Use the Opposite Angles Property**: In a cyclic quadrilateral, the sum of opposite angles is 180°. Therefore, we can find angle BCD:

\[ \angle DAB + \angle BCD = 180° \]

Substituting the value of angle DAB:

\[ 40° + \angle BCD = 180° \]

Solving for angle BCD gives:

\[ \angle BCD = 180° - 40° = 140° \]

5. **Find Angle DBC**: In triangle BCD, since BC = CD, triangle BCD is isosceles. Let angle DBC = angle BDC = θ. Therefore:

\[ \theta + \theta + 140° = 180° \]

Simplifying this gives:

\[ 2\theta + 140° = 180° \]

\[ 2\theta = 180° - 140° = 40° \]

\[ \theta = \frac{40°}{2} = 20° \]

Thus, angle DBC is:

\[ \angle DBC = 20° \]

**Final Answer**: The value of angle DBC is 20°.

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