Two chords AB, CD of a circle with centre O intersect each other at P. `/_ADP = 23^@` and `/_APC = 70^@`, then the `/_BCD `is

Options

A. `45^@`
B. `47^@`
C. `57^@`
D. ` 67^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two chords AB and CD of a circle that intersect at point P. We are given:
- Angle ADP = 23°
- Angle APC = 70°

We need to find the measure of angle BCD.

### Step 2: Use the property of linear pairs
Since angles APC and APD are on a straight line (CD), they form a linear pair. Therefore, we can write:
\[ \angle APC + \angle APD = 180° \]

### Step 3: Calculate angle APD
Substituting the known value of angle APC:
\[ 70° + \angle APD = 180° \]
\[ \angle APD = 180° - 70° = 110° \]

### Step 4: Use the triangle property
Now, in triangle APD, we can find angle PAD using the triangle sum property:
\[ \angle PAD + \angle ADP + \angle APD = 180° \]
Substituting the known values:
\[ \angle PAD + 23° + 110° = 180° \]
\[ \angle PAD + 133° = 180° \]
\[ \angle PAD = 180° - 133° = 47° \]

### Step 5: Use the property of angles in the same segment
According to the property of angles in the same segment of a circle, angles PAD and BCD are equal:
\[ \angle PAD = \angle BCD \]
Thus:
\[ \angle BCD = 47° \]

### Final Answer
The measure of angle BCD is **47°**.

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