Geometry Problem

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

Click the figure below to see the complete problem 665.

## Wednesday, September 7, 2011

### Problem 665: Intersecting Circles, Secant, Tangent, Angles, Mind Map

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http://img33.imageshack.us/img33/974/problem665.png

ReplyDeleteDenote (XYZ)=angle(XYZ)

CO interesect DO’ at F

Note that (ECA)=(ABC)=beta

And (ABD)=(ADE)=alpha

Quadraliteral FCED is cyclic with (CFD)=(ECD)+(EDC)= alpha+beta=(CBD) (see picture)

So 5 points F,C,E,D, B co-cyclic

And ( EBD)=(ECD)= beta ( angles face the same arc DE)

So x= (ABD)-(EBD)= alpha-beta

Peter Tran

Join BC, BD

ReplyDelete∠CBD = ∠ABD + ∠ABC = α + β

∠CED = 180° − (α + β)

∴B, C, E, D are concyclic

So α = ∠CBE = x + ∠ABC = x + β

∴ x = α − β

http://imgsrc.baidu.com/forum/pic/item/38dbb6fd5266d0162ed21d77972bd40734fa35a1.jpg

ReplyDeletechinese

< CBA = beta and < ABD = alpha hence < CBD + CED = 180

ReplyDeleteSo < EBD = beta = alpha - x hence x = alpha - beta

Sumith Peiris

Moratuwa

Sri Lanka