circle geometry (1 Viewer)

[bos1234]

arcD = arcAB
Prove AC = DB and <ADC = <DAB

 

[jyu]

arcCD = arcAB

Prove AC = DB and



arcCD = arcAB
.: CD = AB
angle ADB = angle CAD [subtended by equal arcs on the circumference]angle ABD = angle ACD [subtended by the same arcAD on the circumference]

.: triangles ACD and DBA are congruent.
.: AC = DB


<ADC ="<DAB

</p">

 

[watatank]

I'm not exactly sure what exactly you are trying to ask here, but I can tell you all the information you can get from this question....

I'm going to assume that

arcAB = arc CD

and you want to prove AC = DB

Let where CB and AD meet be point E

Ok ^CAD = ^ACB (angles subtending equal length arcs, arcAB = arc CD)

Therefore CE = AE (Base angles of triangleACE are equal)

^CED = ^AEB (Vertically opposite angles are equal)

therefore triangleCED is congurent to triangle AEB (two angles in one triangle and one side respectively equal to two angles and one side in another....or AAS congurence test )

Since they are congurent, DE = BE (corresponding sides equal)

EDIT: Hmm that sux I think I just did it the very long way