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Geometry ProblemPost your solution in the comment box belowLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 925.
http://img39.imageshack.us/img39/8260/xhe.bmpLet EG cut CF at K. define angles x and y as shown on the sketch In triangle BEK we have y=∠ (BEG)- ∠ (CBD)=1/4(Arc(AB)+Arc(CD))- 1/2Arc(CD)= 1/4(Arc(AB)-Arc(CD))We have x= ½∠ (BFA)=1/2 x ½( Arc(AB)-Arc(CD))= 1/4(Arc(AB)-Arc(CD))So x=y and EG//FH
Let EG and FH cut AB at P,Q respectively∠BFQ = 1/2∠BFA = 1/2(90 - A/2 - B/2)∠BQF = a + ∠BFQ = 90 + A/2 - B/2Now let ∠CBD = x, ∠CDB = y, so that x+y=a.∠PEA = 1/2∠BEA = 1/2∠(180 - y - (B-x)) = 90 - y/2 - B/2 + x/2∠BPE = ∠PEA + y = 90 - B/2 + x/2 + y/2 = 90 - B/2 + A/2So ∠BPE=∠BQFq.e.d.
See following picture:http://img40.imageshack.us/img40/9981/ptg.gif