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Simranjeet Singh , asked a question
Subject: Math , asked on 1/3/13

the bisectors of the opposite angles A and C of cyclic quadrilateral ABCD intersect the circle at the point E and F.prove that EF is a diameter of the circle.

Ayushi Sharma From M.g.m. Sr Sec School, added an answer, on 1/3/13
364 helpful votes in Math

Given: ABCD is a cyclic quadrilateral. AE and CF are the bisectors of ∠A and ∠C respectively.

To prove: EF is the diameter of the circle i.e. ∠EAF = 90°

Construction: Join AE and FD.

Proof:

ABCD is a cyclic quadrilateral.

∴ ∠A + ∠C = 180° (Sum of opposite angles of a cyclic quadrilateral is 180° )

⇒ ∠EAD + ∠DCF = 90° ...(1) (AE and CF are the bisector of ∠A and ∠C respectively)

∠DCF = ∠DAF ...(2) (Angles in the same segment are equal)

From (1) and (2), we have

∠EAD + ∠DAF = 90°

⇒ ∠EAF = 90°

⇒ ∠EAF is the angle in a semi-circle.

⇒ EF is the diameter of the circle.

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Ayushi Sharma From M.g.m. Sr Sec School, added an answer, on 1/3/13
364 helpful votes in Math

Given: ABCD is a cyclic quadrilateral. AE and CF are the bisectors of ∠A and ∠C respectively.

To prove: EF is the diameter of the circle i.e. ∠EAF = 90o

Construction: Join AE and FD.

Proof:

ABCD is a cyclic quadrilateral.

∴ ∠A + ∠C = 180o (Sum of opposite angles of a cyclic quadrilateral is 180o )

⇒ ∠EAD + ∠DCF = 90o ...(1) (AE and CF are the bisector of ∠A and ∠C respectively)

∠DCF = ∠DAF ...(2) (Angles in the same segment are equal)

From (1) and (2), we have

∠EAD + ∠DAF = 90o

⇒ ∠EAF = 90o

⇒ ∠EAF is the angle in a semi-circle.

⇒ EF is the diameter of the circle.

  • Was this answer helpful?
  • 2
100% users found this answer helpful.

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