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
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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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