## Thursday, April 6, 2023

### Geometry Problem 1533: Discovering Relationships between Angles and Lines in an Exterior Right Triangle of a Square

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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1. Ang x = 82°

2. Let EF cut AB at P.

AOBE 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

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

4. https://photos.app.goo.gl/5dpTKYTQDr462XLL7

5. Same reasoning

Let 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