To prove:

In a cyclic quadrilateral, the sum of each pair of opposite angles is 180.

Given

A quadrilateral (ABCD) inscribed in a circle (i.e., a cyclic quadrilateral).

We must prove:

[
angle A + angle C = 180^circ
]
and
[
angle B + angle D = 180^circ
]

Proof

A key property of circles is:

An inscribed angle equals half the measure of its intercepted arc.

Step 1: Express angles using intercepted arcs

In cyclic quadrilateral (ABCD):

• (angle A) subtends arc (BCD)
• (angle C) subtends arc (BAD)

So,

[
angle A = frac{1}{2}(text{arc } BCD)
]

[
angle C = frac{1}{2}(text{arc } BAD)
]

Step 2: Add the two angles

[
angle A + angle C

frac{1}{2}(text{arc } BCD)
+
frac{1}{2}(text{arc } BAD)
]

[

frac{1}{2}(text{arc } BCD + text{arc } BAD)
]

But,

[
text{arc } BCD + text{arc } BAD = 360^circ
]

(since together they make the whole circle)

Step 3: Simplify

[
angle A + angle C

frac{1}{2}(360^circ)

180^circ
]

Similarly,

[
angle B + angle D = 180^circ
]

(by the same reasoning using their intercepted arcs)

Conclusion

In a cyclic quadrilateral,

[
boxed{angle A + angle C = 180^circ}
]

[
boxed{angle B + angle D = 180^circ}
]

Hence, the sum of either pair of opposite angles of a cyclic quadrilateral is 180.

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