Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 958.
Thursday, January 16, 2014
Geometry Problem 958. Isosceles Triangle, Altitude, 45 Degrees, Perpendicular, Angle Bisector, Metric Relations
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Note that G is the center of excircle of triangle BDE
ReplyDeleteRadius of excircle Rd= GH= p-ED .. where p= half of perimeter
2p=ED+EB+BD
p-ED=1/2(EB+BD-ED) … (1)
Since ADE is right isosceles triangle => ED=EA
(1) become GH= ½(EB+BD-EA)=1/2(BD-BA)=1/2(BD-BC)=1/2CD