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

Labels:
45 degrees,
altitude,
angle bisector,
isosceles,
metric relations,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

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