Question

prove that any four vertices of a regular pentagon are concyclic

Answer 1

Answer:

Given ABCDE is a regular pentagon .

let ABCDE is a regular pentagon

we know that ,

Each angle of a regular pentagon = 108 degree

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

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> Given: A regular pentagon ABCDE.

> To prove: Every set of four vertices of ABCDE is a set of points lying on a circle.

> Proof:

✴️ First we show that the points A, B, C, E lie on a circle.

✴️ Join AC and BE.

✴️ In ∆ABC and ∆BAE, we have:

✴️ AB = BA (common)

✴️ BC = AE (sides of a regular pentagon)

✴️ angle ABC = angle BAE (each equal to 108°)

✴️ Therefore, ∆ABC is congruent to ∆BAE (by SAS congruency rule)

✴️ => angle BCA = angle AEB (by C.P.C.T.)

✴️ Thus, AB subtends equal angle at two points C and E on the same side of AB.

✴️ Therefore, the points A, B, C, E are concyclic

✴️ Similarly, every set of four vertices of pentagon ABCDE is a set of concyclic points.

✨……Hence, PROVED……✨

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