Wednesday, August 31, 2016

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

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.

Click the figure below to view more details of problem 1254.


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

2 comments:

  1. Join AE and form the triangle AHE.
    Let 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

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  2. Problem 1254
    Is <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

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