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I'm having some trouble with one particular geometry proof:
From that I've drawn the following:
http://img96.imageshack.us/img96/139/circle9we.gif [Broken]
[tex]\angle ADB = \angle CED[/tex] (as [tex]\angle ADB[/tex] and [tex]\angle CED[/tex] are alternant sements)
[tex]\angle CBD = 180 - \angle CED [/tex] (1) (as they are opposite angles in a cyclic quadrilateral)
[tex]180 - \angle CBD = \angle DBA[/tex] (2) as CBA is a straight line
[tex]\therefore \angle CED = \angle DBA[/tex] substitute (1) into (2) This indicates that [tex]\angle BDA = \angle DBA[/tex]. Is that right? And is that enough to prove that [tex]\bigtriangleup ABD \sim \bigtriangleup CDE[/tex] ?
From a point A outside a circle, a secant ABC is drawn cutting the circle at B and C, and a tangent AD touching at D. A chord DE is drawn equal in length to chord DB. Prove that triangles ABD and CDE are similar.
From that I've drawn the following:
http://img96.imageshack.us/img96/139/circle9we.gif [Broken]
[tex]\angle ADB = \angle CED[/tex] (as [tex]\angle ADB[/tex] and [tex]\angle CED[/tex] are alternant sements)
[tex]\angle CBD = 180 - \angle CED [/tex] (1) (as they are opposite angles in a cyclic quadrilateral)
[tex]180 - \angle CBD = \angle DBA[/tex] (2) as CBA is a straight line
[tex]\therefore \angle CED = \angle DBA[/tex] substitute (1) into (2) This indicates that [tex]\angle BDA = \angle DBA[/tex]. Is that right? And is that enough to prove that [tex]\bigtriangleup ABD \sim \bigtriangleup CDE[/tex] ?
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