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 see the complete problem 1005.
Let BD cut circle O at FDraw altitude BH of triangle ABC F is the midpoint of arc AC => OF ⊥AC∠(OFB)= ∠(FBH)= ∠(BEH)Since ∆(OBF) and ∆(BQE) are isosceles => ∠(OBF)= ∠(QBE)So ∠(OBQ)=90 degrees
<DBE=90, and <EBQ=90-(<A+<B/2)=B/2+90-(<A+<B)=<ABD+<OBA=<OBD, so <OBQ=<DBE
Let be B,F, intersection points of circles. FQB isoceles, QO altitude => QO bisector => QB tg to OB
Since QD = QB, < QDB = < QBD so < QBC = < BACHence QB is a tangent to circle ABC at B and so OB has to be perpendicular to QBSumith PeirisMoratuwaSri Lanka
Problem 1005Solution 1In triangle ABC by the theorem of the internal and external dichotomy we haveAB/BC=AD/DC=AE/EC. Is <DBE=90 so the Q is midpoint DE. From theorem Newton apply QD^2=QC.QA=QB^2 so QB is tangent meets the circle with center O.Solution 2<CBQ=<CBE-<QBE=(<BAC+<ACB)/2-<QEB=(90-<ABC/2)-(90-<QDB)=<QDB-<ABC/2=<BAC+<ABC/2-<ABC/2=<BAC. So QB is tangent meets the circle with center O.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE