tag:blogger.com,1999:blog-6933544261975483399.post5353171521763118111..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1005: Triangle, Circle, Circumcenter, Interior and Exterior Angle Bisector, 90 Degree, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-6914583947463772212016-09-03T06:23:22.476-07:002016-09-03T06:23:22.476-07:00Problem 1005
Solution 1
In triangle ABC by the t...Problem 1005<br />Solution 1<br />In triangle ABC by the theorem of the internal and external dichotomy we have<br />AB/BC=AD/DC=AE/EC. Is <DBE=90 so the Q is midpoint DE. From theorem <br /> Newton apply QD^2=QC.QA=QB^2 so QB is tangent meets the circle with center O.<br />Solution 2<br /><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.<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55866997376794639612016-04-09T04:34:24.594-07:002016-04-09T04:34:24.594-07:00Since QD = QB, < QDB = < QBD so < QBC = &...Since QD = QB, < QDB = < QBD so < QBC = < BAC<br /><br />Hence QB is a tangent to circle ABC at B and so OB has to be perpendicular to QB<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45149791467437219942016-03-22T09:07:10.338-07:002016-03-22T09:07:10.338-07:00Let be B,F, intersection points of circles.
FQB ...Let be B,F, intersection points of circles. <br />FQB isoceles, QO altitude => QO bisector => QB tg to OBc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65100479066763377992014-04-17T15:48:53.533-07:002014-04-17T15:48:53.533-07:00<DBE=90, and <EBQ=90-(<A+<B/2)=B/2+90-...<DBE=90, and <EBQ=90-(<A+<B/2)=B/2+90-(<A+<B)=<ABD+<OBA=<OBD, so <OBQ=<DBEIvan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11635839673315659422014-04-16T16:40:55.105-07:002014-04-16T16:40:55.105-07:00Let BD cut circle O at F
Draw altitude BH of trian...Let BD cut circle O at F<br />Draw altitude BH of triangle ABC<br /> F is the midpoint of arc AC => OF ⊥AC<br />∠(OFB)= ∠(FBH)= ∠(BEH)<br />Since ∆(OBF) and ∆(BQE) are isosceles => ∠(OBF)= ∠(QBE)<br />So ∠(OBQ)=90 degrees<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com