Question

Prove that any four vertices of a regular pentagon lies on a circle.

Answer 1

Given ABCDE is a regular pentagon
That is AB = BC = CD = DE = AE
Recall that the sum of angles in a regular pentagon is 540°
Hence each of the interior angle is (540°/5) = 108°
In ΔADE, AE = DE
∴ ∠ADE = ∠DAE [Angles opposite to equal sides are equal]
∠ADE + ∠DAE +∠AED = 180°
∠ADE + ∠ADE + 108° = 180°
2∠ADE = 72°
∴ ∠ADE = 36°
∠ADE = ∠DAE = 36°
⇒ ∠DAB = 108° – 36° = 72°
Consider the quadrilateral ABCD
∠DAB + ∠C = 72° + 108°
That is ∠DAB + ∠C = 180°
Since the sum of the opposite angles of a quadrilateral is supplementary, quadrilateral ABCDE is a cyclic quadrilateral

Answer 2

Hi!

Here is the answer to your question.

Let ABCDE a regular pentagon

It is required to show ABCD is a cyclic quadrilateral.


In ΔAED, AE = ED (Sides of regular pentagon ABCDE)

∴ ∠EAD = ∠EDA

In ΔAED, ∠AED + ∠EAD + ∠EDA = 180°

⇒ 108° + ∠EAD + ∠EAD = 180°

⇒ 2∠EAD = 180° – 108° = 72°

⇒ ∠EAD = 36°

∴ ∠EDA = 36°

∠BAD = ∠BAE – ∠EAD = 108° – 36° = 72°

In quadrilateral ABCD, ∠BAD + ∠BCD = 108° + 72° = 180°

∴ABCD is a cyclic quadrilateral.

Hope it helps.

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