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

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

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## Wednesday, August 31, 2016

### Geometry Problem 1254: Circle, Arc, Chord, Midpoint, Cyclic Quadrilateral, Concyclic Point

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Join AE and form the triangle AHE.

ReplyDeleteLet Angle BAD = A => Angle DAE = Angle HAE = A (since D is midpoint of B & E)

Let Angle ABC = B => Angle CEA = Angle HEA = B => Angle AHE = Angle CHG = 180-(A+B)

and since C is midpoint of AB => Angle CAB = B and triangle ABC is isosceles

=> CF perpendicular to AB and ABG is isosceles => Angle ABG = A

Therefore Angle CBG = A+B & Angle CHG = 180-(A+B) => BCGH are concyclic

Problem 1254

ReplyDeleteIs <GBA=<GAB=<DAB (AF=FB, FG=perpendicular bisector of AB),<BAD=arcBD=arcDE.

Now <CHG+<CBG=(arcCB+arcBG+arcAE)/2+arcCA/2+<BAD=

(arcCB+arcBG+arcAE+arcCA+arcDE)/2=360/2=180.

Therefore CBGH is cyclic. 3

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE