Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to enlarge the problem 940.
Angle ABC =angle BFC =angle AHCSo ABHC concylic and since ABCO concylic by tangent property , we have ABHCO concylicAngle OHC = angle OCA = 90The second part is by the property of chord, straight forwardQED
Notice that BDCE is a harmonic quadrilateral, so the cross ratio of pencil (FE, FD ; FC, FB) = -1. So, if we intersect this pencil with line EA, we will obtain a set of harmonic points, but FB cuts EA at the point of infinity, thus (E, D ; H, P∞) = -1, and it follows that H is the midpoint of DE and it follows that OH ⊥ DE.Proved.
http://img21.imageshack.us/img21/2474/uim3.pngSince FB//ED => Arc(EF)=Arc(BD)We have ∠ (CHD)= ½(Arc(EF)+ Arc(DC)) = ½(Arc(BD)+ Arc(DC))= 1/2Arc(BC)= ∠ (COA)So quadrilateral OCAH is cyclic => OH ⊥ HA and H is the midpoint of ED
Call <A=a. <DBA+<BAD=<EDF+<FDB, but <EDF=<DBA so <BAD=<FDB=b. <OCH=90-(b+90-a/2)=a/2-b=<OAH which makes HACO cyclic.