Question

An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.

Given: ABCD is cyclic,

`square` is the exterior angle of ABCD

To prove: ∠DCE ≅ ∠BAD

Proof: `square` + ∠BCD = `square`   ...[Angles in linear pair] (i)

ABCD is a cyclic.

`square` + ∠BAD = `square`   ...[Theorem of cyclic quadrilateral] (ii)

By (i) and (ii)

∠DCE + ∠BCD = `square` + ∠BAD

∠DCE ≅ ∠BAD

Answer 1

Proof: 

Given: ABCD is cyclic,

\[\boxed{\text{∠DCE}}\] is the exterior angle of ABCD

To prove: ∠DCE ≅ ∠BAD

Proof: \[\boxed{\text{∠DCE}}\] + ∠BCD = \[\boxed{180°}\]   ...[Angles in linear pair] (i)

ABCD is a cyclic.

\[\boxed{\text{∠BCD}}\] + ∠BAD = \[\boxed{180°}\]   ...[Theorem of cyclic quadrilateral] (II)

By (i) and (ii)

∠DCE + ∠BCD = \[\boxed{\text{∠BCD}}\] + ∠BAD

∠DCE ≅ ∠BAD