prove 4 vertices of any regular pentagon are concyclic
Answers
Answered by
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.
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.
RADHIKAA26:
thanks
pictur please
sorry i don't send picture
i give instruction
draw a Pentagon ABCDE join AD it makes 108° at angle DEA
Answered by
3
Step-by-step explanation:
hope it helps ✌️✌️✌️✌️✌️✌️✌️✌️✌️✌️❣️❣️❣️❣️❣️❣️❣️
Attachments:
Similar questions