ABCD is a cyclic quadrilateral whose diagonals intersect at P. If `AB = BC, /_DBC = 70^@ and /_BAC = 30^@`, then the measure of `/_PCD` is :

Options

A. `35^@`
B. `50^@`
C. `55^@`
D. `30^@`

Answer

To solve the problem, we need to find the measure of angle \( \angle PCD \) in the cyclic quadrilateral \( ABCD \) with the given conditions. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- We have a cyclic quadrilateral \( ABCD \).
- The diagonals \( AC \) and \( BD \) intersect at point \( P \).
- Given:
- \( AB = BC \) (which means triangle \( ABC \) is isosceles)
- \( \angle DBC = 70^\circ \)
- \( \angle BAC = 30^\circ \)

### Step 2: Find \( \angle BAD \)
Since \( AB = BC \), we know that:
- \( \angle BAC = \angle BCA \)

Thus, we can denote:
- \( \angle BCA = 30^\circ \)

Now, we can find \( \angle ABC \):
- In triangle \( ABC \):
\[ \angle ABC + \angle BAC + \angle BCA = 180^\circ \]
\[ \angle ABC + 30^\circ + 30^\circ = 180^\circ \]
\[ \angle ABC = 180^\circ - 60^\circ = 120^\circ \]

### Step 3: Find \( \angle BAD \)
Next, we can find \( \angle BAD \):
- Since \( \angle DBC = 70^\circ \):
- By the property of angles in the same segment:
\[ \angle CBD = \angle CAD = 70^\circ \]
- Therefore, we can find \( \angle BAD \):
\[ \angle BAD = \angle BAC + \angle CAD = 30^\circ + 70^\circ = 100^\circ \]

### Step 4: Use the Property of Cyclic Quadrilaterals
In a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \):
\[ \angle BAD + \angle BCD = 180^\circ \]
Substituting \( \angle BAD = 100^\circ \):
\[ 100^\circ + \angle BCD = 180^\circ \]
\[ \angle BCD = 180^\circ - 100^\circ = 80^\circ \]

### Step 5: Find \( \angle PCD \)
Now, we know that:
- \( \angle BCD = 80^\circ \)

Since \( \angle BCD \) and \( \angle PCD \) are angles on the same line (formed by the intersection of diagonals), we can find \( \angle PCD \):
- In triangle \( BCD \):
\[ \angle BCD + \angle PCD = 180^\circ \]
Thus:
\[ \angle PCD = 180^\circ - 80^\circ = 100^\circ \]

### Final Answer
The measure of \( \angle PCD \) is \( 100^\circ \).

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