ABCD is cyclic quadrilateral. Sides AB and DC, when produced meet at the point P and sides AD and BC, when produced meet at the point Q. If `angle ADC = 85^(@) and angle BPC = 40^(@)`, then `angle CQD` is equal to

Options

A. `30^(@)`
B. `40^(@)`
C. `55^(@)`
D. `85^(@)`

Answer

To solve the problem, we need to find the measure of angle CQD in the cyclic quadrilateral ABCD, given the angles ADC = 85° and BPC = 40°.

### Step-by-Step Solution:

1. **Understand the Properties of Cyclic Quadrilaterals**:
In a cyclic quadrilateral, the sum of the opposite angles is supplementary. That means:
\[ \angle A + \angle C = 180^\circ \quad \text{and} \quad \angle B + \angle D = 180^\circ \]

2. **Identify the Angles**:
We know:
- \(\angle ADC = 85^\circ\)
- \(\angle BPC = 40^\circ\)

3. **Find Angle CDQ**:
Since ADQ is a straight line, we can use the linear angle property:
\[ \angle ADC + \angle CDQ = 180^\circ \]
Substituting the known value:
\[ 85^\circ + \angle CDQ = 180^\circ \]
Therefore:
\[ \angle CDQ = 180^\circ - 85^\circ = 95^\circ \]

4. **Find Angle ABC**:
Again using the property of cyclic quadrilaterals:
\[ \angle ABC + \angle ADC = 180^\circ \]
Thus:
\[ \angle ABC = 180^\circ - 85^\circ = 95^\circ \]

5. **Find Angle PCB**:
In triangle PBC, we apply the angle sum property:
\[ \angle BPC + \angle ABC + \angle PCB = 180^\circ \]
Substituting the known values:
\[ 40^\circ + 95^\circ + \angle PCB = 180^\circ \]
Therefore:
\[ \angle PCB = 180^\circ - 135^\circ = 45^\circ \]

6. **Find Angle QDC**:
Using the linear angle property again:
\[ \angle BPC + \angle QDC = 180^\circ \]
Thus:
\[ 40^\circ + \angle QDC = 180^\circ \]
Therefore:
\[ \angle QDC = 180^\circ - 40^\circ = 140^\circ \]

7. **Find Angle CQT**:
In triangle CDQ, we apply the angle sum property:
\[ \angle CQD + \angle QDC + \angle QCD = 180^\circ \]
We already found:
- \(\angle QDC = 95^\circ\)
- \(\angle QCD = 45^\circ\)
Thus:
\[ \angle CQD + 95^\circ + 45^\circ = 180^\circ \]
Therefore:
\[ \angle CQD = 180^\circ - 140^\circ = 40^\circ \]

### Final Answer:
\(\angle CQD = 40^\circ\)