In the given figure,AB is parallel to DC. ` angle BCE = 80 ^(@) and angle BAC = 25^(@). Find: (i) ` angle CAD `
(ii) ` angle CBD `
(iii) ` angle ADC `

Answer

To solve the problem step by step, we will use the properties of parallel lines and angles in cyclic quadrilaterals.

### Given:
- AB is parallel to DC
- ∠BCE = 80°
- ∠BAC = 25°

### To find:
(i) ∠CAD
(ii) ∠CBD
(iii) ∠ADC

### Step-by-step Solution:

**Step 1: Find ∠CAD**

Since AB is parallel to DC, we can use the property of alternate interior angles. The exterior angle ∠BCE is equal to the interior opposite angle ∠BAD.

Thus, we have:
\[ \angle BAD = \angle BCE = 80° \]

Now, we can find ∠CAD using the relationship:
\[ \angle CAD = \angle BAD - \angle BAC \]
Substituting the known values:
\[ \angle CAD = 80° - 25° = 55° \]

**Step 2: Find ∠CBD**

Since angles CAD and CBD subtend the same arc AC in the cyclic quadrilateral ABCD, they are equal:
\[ \angle CBD = \angle CAD = 55° \]

**Step 3: Find ∠ADC**

In the isosceles trapezium ABCD, the angles ∠ADC and ∠BCD are equal because they are angles on the same segment. We can find ∠ADC using the fact that the sum of angles in a triangle is 180°.

We know:
\[ \angle BCD = \angle BCE = 80° \]
Thus,
\[ \angle ADC + \angle BCD = 180° \]
Substituting the known value:
\[ \angle ADC + 80° = 180° \]
Solving for ∠ADC:
\[ \angle ADC = 180° - 80° = 100° \]

### Final Answers:
(i) ∠CAD = 55°
(ii) ∠CBD = 55°
(iii) ∠ADC = 100°