Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 906.

## Saturday, August 3, 2013

### Problem 906: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Perpendicular

Labels:
bicentric quadrilateral,
chord,
circle,
circumcircle,
incircle,
perpendicular

Subscribe to:
Post Comments (Atom)

A=a

ReplyDeleteC=180-a

FHA=BFH=b [=FGH]

AEG=EGD=c [=EFG]

FH meet EG at K

EKH=360-a-b-c=FKG

CGK=180-EGD=180-c

CFK=180-b

360=180-a+180-b+180-c+360-a-b-c

a+b+c=270

FKE=90

Let EG and FH meet at K.

ReplyDeleteThen

∠FKG

= 1/2 ( ∠FIG + ∠EIH )

= 1/2 ((180 - ∠FCG ) + (180 - ∠EAH))

= 90 °