Geometry Problem

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

Click the figure below to see the complete problem 910.

## Friday, August 9, 2013

### Problem 910: Bicentric Quadrilateral, Distance between the Incenter and Circumcenter, Incircle, Circumcircle, Circumscribed, Inscribed, Circumradius

Labels:
bicentric quadrilateral,
circumcenter,
circumradius,
distance,
incenter

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∠KDJ

ReplyDelete= ∠KDA + ∠ADJ

= ∠KDA + ∠ABJ

= 1/2 [∠CDA + ∠ABC]

= 90°

⇒ KJ is diameter of the circle O

⇒ OK = OJ = R

In ΔOIK,

IK² = R² + d² - 2Rd cos∠IOK

In ΔOIJ,

IJ² = R² + d² - 2Rd cos∠IOJ

But ∠IOK + ∠IOJ = 180°,

cos∠IOK = −cos∠IOJ

Adding up, we have

IK² + IJ² = 2 (R² + d²)

KI^2=R^2+d^2-2*R*d*cos<KOI, and IJ^2=R^2+d^2+2*R*d*cos<KOI. Adding the two equations gives the result.

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