Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, August 21, 2016
Geometry Problem 1251: Triangle, Circle, Diameter, Perpendicular, 90 Degrees
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https://goo.gl/photos/4guoy5kN3aqGP59PA
ReplyDeleteDraw circle O1 diameter GB and circle O2 diameter HB
Note that circle O1 pass through points F and D
Circle O2 pass through points F and E
We have ∠( GBF)= ∠(GDF)= x
∠(FBH)= ∠(FEH)=y
But ∠(FEC)= ∠(FDC)=y
But ∠(GDF)+ ∠(FDC)=90= x+y
So ∠ (GBH)= ∠(GBF)+ ∠(FBH)=x+y=90
Note that GBDF, AEDF and BEFH are all cyclic quadrilaterals.
ReplyDeleteHence < GBF = < GDF = < AEF = < BHF
So < GBH = < GBF + < HBF = < BHF + <HBF = 90
Sumith Peiris
Moratuwa
Sri Lanka