In `Delta ABC`, side AB is produced to D so that BD = BC. If `angleB= 60^(@) and angleA= 70^(@)` then

Options

A. `AD gt CD`
B. `angleADC= 90^(@)`
C. `AD lt CD`
D. `angleCAD= 30^(@)`

Answer

To solve the problem step by step, we will analyze the triangle ABC and the points defined in the question.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Triangle ABC with angles:
- \( \angle A = 70^\circ \)
- \( \angle B = 60^\circ \)
- Side AB is extended to point D such that \( BD = BC \).

2. **Calculate Angle C:**
- The sum of angles in a triangle is \( 180^\circ \).
- Therefore, we can find \( \angle C \):
\[ \angle C = 180^\circ - \angle A - \angle B = 180^\circ - 70^\circ - 60^\circ = 50^\circ \]

3. **Determine Angle CBD:**
- Since \( BD = BC \), triangle BCD is isosceles.
- The exterior angle \( \angle CBD \) can be calculated as:
\[ \angle CBD = \angle A + \angle C = 70^\circ + 50^\circ = 120^\circ \]

4. **Find Angles BDC and BCD:**
- In triangle BCD, the angles must sum to \( 180^\circ \):
\[ \angle BDC + \angle BCD + \angle CBD = 180^\circ \]
- Substituting \( \angle CBD = 120^\circ \):
\[ \angle BDC + \angle BCD + 120^\circ = 180^\circ \]
- This simplifies to:
\[ \angle BDC + \angle BCD = 60^\circ \]
- Since \( \angle BDC = \angle BCD \) (because \( BD = BC \)):
\[ 2 \angle BDC = 60^\circ \implies \angle BDC = 30^\circ \quad \text{and} \quad \angle BCD = 30^\circ \]

5. **Analyze Triangle ADC:**
- In triangle ADC:
- \( \angle DAC = 70^\circ \)
- \( \angle DCA = \angle C + \angle BDC = 50^\circ + 30^\circ = 80^\circ \)

6. **Compare Angles and Sides:**
- Since \( \angle DCA = 80^\circ \) is greater than \( \angle DAC = 70^\circ \), by the property of triangles, the side opposite to the greater angle is longer.
- Thus, side \( AD \) (opposite \( \angle DCA \)) is greater than side \( CD \) (opposite \( \angle DAC \)):
\[ AD > CD \]

7. **Conclusion:**
- Therefore, the correct option is that \( AD \) is greater than \( CD \).