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.

## Wednesday, April 16, 2014

### Geometry Problem 1005: Triangle, Circle, Circumcenter, Interior and Exterior Angle Bisector, 90 Degree, Perpendicular

Labels:
90,
angle bisector,
circle,
circumcenter,
perpendicular,
triangle

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Let BD cut circle O at F

ReplyDeleteDraw 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

ReplyDeleteLet be B,F, intersection points of circles.

ReplyDeleteFQB isoceles, QO altitude => QO bisector => QB tg to OB

Since QD = QB, < QDB = < QBD so < QBC = < BAC

ReplyDeleteHence QB is a tangent to circle ABC at B and so OB has to be perpendicular to QB

Sumith Peiris

Moratuwa

Sri Lanka

Problem 1005

ReplyDeleteSolution 1

In triangle ABC by the theorem of the internal and external dichotomy we have

AB/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