Geometry Problem

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

Click the figure below to see the complete problem 623.

## Thursday, June 16, 2011

### Problem 623: Triangle, Incenter, Excenter, Incircle, Excircle, Concyclic Points

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∠IBE = B/2 + (180°-B)/2 = 90°

ReplyDeleteSimilarly ∠ICE = 90°

So ∠IBE + ∠ICE = 180° and I,B,E,C are concyclic

internal and external angle bisector are perpendicular; so the result follows

ReplyDelete< ECB = 90-C/2 = A/2 + B/2

ReplyDeleteSo < AEC = B/2 = < IBC and the result follows

Sumith Peiris

Moratuwa

Sri Lanka