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