The side AC of a `Delta ABC` is extended to D such that BC = CD. If `angleACB` is `70^@` ,then what is `angle ADB` equal to ?

Options

A. `35^@`
B. `45^@`
C. `70^@`
D. `110^@`

Answer

To find the measure of angle ADB in triangle ABC, where side AC is extended to point D such that BC = CD and angle ACB = 70°, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- We know that angle ACB = 70°.
- BC = CD (the lengths of segments BC and CD are equal).

2. **Understand the Geometry:**
- Since AC is extended to D, angle ACD is formed.
- Angle ACD can be expressed as the sum of angle ACB and angle BCD (which we will find).

3. **Use the Straight Line Property:**
- Since ACD is a straight line, we have:
\[ \text{angle ACB} + \text{angle BCD} = 180° \]
- Therefore:
\[ 70° + \text{angle BCD} = 180° \]
- From this, we can find angle BCD:
\[ \text{angle BCD} = 180° - 70° = 110° \]

4. **Apply the Isosceles Triangle Property:**
- Since BC = CD, triangle BCD is isosceles. Thus, the angles opposite the equal sides are equal:
\[ \text{angle CDB} = \text{angle CBD} = \theta \]

5. **Set Up the Equation for Triangle BCD:**
- The sum of angles in triangle BCD is:
\[ \text{angle BCD} + \text{angle CDB} + \text{angle CBD} = 180° \]
- Substituting the known values:
\[ 110° + \theta + \theta = 180° \]
- This simplifies to:
\[ 110° + 2\theta = 180° \]

6. **Solve for θ:**
- Rearranging gives:
\[ 2\theta = 180° - 110° = 70° \]
- Dividing by 2:
\[ \theta = \frac{70°}{2} = 35° \]

7. **Conclusion:**
- Since θ represents angle ADB:
\[ \text{angle ADB} = 35° \]

### Final Answer:
Angle ADB is equal to 35°.

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