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