Question

prove that any four vertices of a regular pentagon are concyclic​

Answer 1

Answer:

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.

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Answer 2

Step-by-step explanation:

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