Geometry Problem

Click the figure below to see the complete problem 482 about Triangle, Circumcircle, Incenter, Excenter, Midpoint, Cyclic Points.

See also:

Complete Problem 482

Level: High School, SAT Prep, College geometry

## Tuesday, July 27, 2010

### Problem 482: Triangle, Circumcircle, Incenter, Excenter, Midpoint, Cyclic Points

Labels:
circumcircle,
cyclic quadrilateral,
excenter,
incenter,
midpoint,
triangle

Subscribe to:
Post Comments (Atom)

Note that Angle DCE= Angle DBE=90

ReplyDeleteQuadrilateral DBEC is cyclic with DE is the diameter of circle DBEC

Since F is the midpoint of arc BC , F will locate on the perpendicular bisector of chord BC.

F is the intersecting point of diameter DE and perpen. bisector of chord BC so F must be the center of circle DBEC

Peter Tran

Peter: Why DCE = 90

ReplyDeleteA,D,F,E lie on the line AE (D & E on bisector,F midpo)

BF = FC = FE (1)

( arc BF = arc FC & BF = FE from P156 )

▲DFC isoceles, ang FCD = C/2 + A/2 = 90 - B/2

ang DFC = ang B (arc AC)

=> ang FDC =180-B-(90-B/2)=90 - B/2

=> ang FDC = ang FCD

=>FD = FC (2)

from (1) & (2)

BF = FC = FE = FD

CD is angle bisector of (ACB)

ReplyDeleteCE is angle bisector of (BCx)

(ACB)+(BCx)=180

so (DCE)=90

Peter Tran

A,D,F,E are collinear points

ReplyDelete< FDB = < FBD = (A+B)/2 hence FB = FD = FC

But < AFB = C and < FBE = 90-B/2 - A/2 = C/2 so < FEB also = C/2 hence FE = FB

SoFD = FB = FE = FC and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Here is an GeoGebra applete on it: http://www.xente.mundo-r.com/ilarrosa/GeoGebra/Circ_Inc_Exinc.html

ReplyDelete