In cyclic quadrilateral `ABCD`, Diagonals `AC` and `BD` intersect at `P`. If `/_DBC = 70^(@)` and `/_BAC = 30^(@)` then find `/_BCD`

Answer

To solve the problem, we need to find the angle \( \angle BCD \) in the cyclic quadrilateral \( ABCD \) where the diagonals \( AC \) and \( BD \) intersect at point \( P \). Given that \( \angle DBC = 70^\circ \) and \( \angle BAC = 30^\circ \), we can follow these steps:

### Step 1: Identify the angles in triangles
We know that in cyclic quadrilaterals, the angles subtended by the same chord are equal. Here, we have two triangles: \( \triangle ABC \) and \( \triangle DBC \).

### Step 2: Use the property of angles in triangles
From the cyclic nature of the quadrilateral, we can state that:
\[ \angle BAC = \angle BDC \]
Thus, we have:
\[ \angle BDC = 30^\circ \]

### Step 3: Sum of angles in triangle \( BCD \)
Now, we can apply the triangle angle sum property in triangle \( BCD \):
\[ \angle BDC + \angle DBC + \angle BCD = 180^\circ \]
Substituting the known values:
\[ 30^\circ + 70^\circ + \angle BCD = 180^\circ \]

### Step 4: Solve for \( \angle BCD \)
Now, we can simplify the equation:
\[ 100^\circ + \angle BCD = 180^\circ \]
Subtracting \( 100^\circ \) from both sides gives us:
\[ \angle BCD = 180^\circ - 100^\circ = 80^\circ \]

### Final Answer
Thus, the measure of angle \( \angle BCD \) is:
\[ \angle BCD = 80^\circ \]

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