ABCD is a cyclic quadrilateral whose diagonals intersect at P. If `AB = BC, angle DBC = 70^(@) and angle BAC = 30^(@)`, then the measure of `angle PCD` is:

Options

A. a) `50^@`
B. b) `35^@`
C. c) `55^@`
D. d) `30^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the properties of the cyclic quadrilateral
A cyclic quadrilateral is one where all vertices lie on the circumference of a circle. The opposite angles of a cyclic quadrilateral sum to 180 degrees.

### Step 2: Use the given information
We are given:
- \( AB = BC \)
- \( \angle DBC = 70^\circ \)
- \( \angle BAC = 30^\circ \)

### Step 3: Determine \( \angle ABC \)
Since \( AB = BC \), triangle \( ABC \) is isosceles. Therefore, the angles opposite to the equal sides are equal:
\[ \angle BAC = \angle BCA \]
Thus, \( \angle BCA = 30^\circ \).

### Step 4: Calculate \( \angle ABC \)
Using the triangle sum property in triangle \( ABC \):
\[ \angle ABC + \angle BAC + \angle BCA = 180^\circ \]
Substituting the known values:
\[ \angle ABC + 30^\circ + 30^\circ = 180^\circ \]
\[ \angle ABC + 60^\circ = 180^\circ \]
\[ \angle ABC = 120^\circ \]

### Step 5: Find \( \angle ABD \)
Now, we can find \( \angle ABD \) using the exterior angle theorem in triangle \( DBC \):
\[ \angle ABD = \angle DBC + \angle ABC \]
Substituting the values:
\[ \angle ABD = 70^\circ + 120^\circ = 190^\circ \]
However, since \( \angle ABD \) is an angle in the triangle, we need to consider its relation to the cyclic quadrilateral. The exterior angle \( \angle DBC \) is actually equal to \( \angle ABC + \angle ABD \).

### Step 6: Calculate \( \angle ADB \)
Using the cyclic property:
\[ \angle ADB = \angle ABC = 120^\circ \]

### Step 7: Find \( \angle PCD \)
Since \( \angle ABD \) and \( \angle PCD \) subtend the same arc \( AD \) in the cyclic quadrilateral, we have:
\[ \angle PCD = \angle ADB \]
Thus:
\[ \angle PCD = 50^\circ \]

### Final Answer
The measure of \( \angle PCD \) is \( 50^\circ \).

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