Geometry Problem 1533. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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Ang x = 82°

ReplyDeleteLet EF cut AB at P.

ReplyDeleteAOBE is a cyclic quadrilateral with AB as diameter.

So since AO=OB, OE bisects <AEB

So OFC = < BPE = 180 - 45 - 53 = 82.

Sumith Peiris

Moratuwa

Sri Lanka

Hi ! This problem is original, is simple , but you decide to display for solution if you think it is appropriate, Thanks

ReplyDeletehttps://photos.app.goo.gl/5dpTKYTQDr462XLL7

ReplyDeleteSame reasoning

ReplyDeleteLet AB ∩ EO = K

∠AOB = 90° (1)

∠AEB = 90° (2)

(1)(2)

⇒ AEBO = cyclic quadrilateral ØAB

⇒ ∠AOK = ∠ABE = 53° (3)

△OAB :

OA = OB

⇒ △OAB isosceles and rectangle at O

⇒ ∠BAO = ∠AOB = 45° (4)

△AKO

(3)(4)

⇒ ∠AKO = 180° - (45° + 53°)

= 82°

= ∠OFC, the dolution