Prove in a cylic - trapezium the non - parallel sides are equal and the diagonals are also equal .

Answer

To prove that in a cyclic trapezium the non-parallel sides are equal and the diagonals are also equal, we can follow these steps:

### Step 1: Draw a cyclic trapezium
Let’s consider a cyclic trapezium \( ABCD \) where \( AB \parallel CD \). This means that the sides \( AB \) and \( CD \) are parallel.

### Step 2: Identify angles
Since \( AB \parallel CD \), we can use the property of co-interior angles. Thus, we have:
\[ \angle B + \angle D = 180^\circ \quad \text{(co-interior angles)} \]
\[ \angle A + \angle C = 180^\circ \quad \text{(co-interior angles)} \]

### Step 3: Set up equations
From the above, we can write:
\[ \angle B + \angle D = \angle B + \angle C \]
This implies:
\[ \angle D = \angle C \]

### Step 4: Draw the diagonals
Now, draw the diagonals \( AC \) and \( BD \). This divides the trapezium into two triangles: \( \triangle ACD \) and \( \triangle BCD \).

### Step 5: Use congruence criteria
In triangles \( ACD \) and \( BCD \):
- \( DC = DC \) (common side)
- \( \angle D = \angle C \) (from Step 3)
- \( \angle A = \angle B \) (angles subtended by the same chord in the same segment)

By the Angle-Side-Angle (ASA) congruence criterion, we can conclude:
\[ \triangle ACD \cong \triangle BCD \]

### Step 6: Conclude equal sides and diagonals
From the congruence of triangles, we have:
- \( AD = BC \) (non-parallel sides are equal)
- \( AC = BD \) (diagonals are equal)

Thus, we have proved that in a cyclic trapezium, the non-parallel sides are equal and the diagonals are also equal.