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