ABCD is a cyclic quadrilateral having `∠ADC = 78°`, where DC and AB are extended so that they meet at E such that `∠CEB` is `52°` and AD and BC extended to meet at F. Find the value of ∠AFB

Options

A. `38^(@)`
B. `78^(@)`
C. `28^(@)`
D. `102^(@)`

Answer

To solve the problem step by step, we will analyze the cyclic quadrilateral ABCD and the angles involved.

### Step 1: Understand the given information
We have a cyclic quadrilateral ABCD with:
- \( \angle ADC = 78^\circ \)
- \( \angle CEB = 52^\circ \)

We need to find \( \angle AFB \) where AD and BC are extended to meet at point F.

### Step 2: Draw the diagram
Draw the cyclic quadrilateral ABCD with points A, B, C, and D. Extend lines DC and AB to meet at point E, and extend lines AD and BC to meet at point F.

### Step 3: Find \( \angle DAE \)
In triangle AED, we know:
- \( \angle AED = \angle CEB = 52^\circ \)
- \( \angle ADC = 78^\circ \)

Using the triangle angle sum property:
\[ \angle AED + \angle DEA + \angle DAE = 180^\circ \]
Substituting the known values:
\[ 52^\circ + 78^\circ + \angle DAE = 180^\circ \]
\[ \angle DAE = 180^\circ - 130^\circ = 50^\circ \]

### Step 4: Find \( \angle ABC \)
In a cyclic quadrilateral, opposite angles sum to \( 180^\circ \):
\[ \angle ADC + \angle ABC = 180^\circ \]
Substituting the known value:
\[ 78^\circ + \angle ABC = 180^\circ \]
\[ \angle ABC = 180^\circ - 78^\circ = 102^\circ \]

### Step 5: Find \( \angle AFB \)
In triangle AFB, we again use the triangle angle sum property:
\[ \angle AFB + \angle ABC + \angle FAB = 180^\circ \]
We already found:
- \( \angle ABC = 102^\circ \)
- \( \angle DAE = \angle FAB = 50^\circ \)

Substituting these values:
\[ \angle AFB + 102^\circ + 50^\circ = 180^\circ \]
\[ \angle AFB + 152^\circ = 180^\circ \]
\[ \angle AFB = 180^\circ - 152^\circ = 28^\circ \]

### Final Answer
Thus, the value of \( \angle AFB \) is \( 28^\circ \).

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