In the attached figure, AB//OC, <ABC=130, Find x,y.
Exactly, Wilmer,
we can join O to B, creating 3 internal isosceles triangles.
We can label identical angles and discover that angles COD and BAO are 80 degrees and BCO is 50 degrees.
This means the angle at B cannot form a parallelogram.
The question is, where do we put the point at x ?
No...
Well...i tried your method...check out this-
ABC+ADC=180 -->[As ABCD is a cyclic quadrilateral so opp <s are
supplementary]
=> 130+ADC=180
=> ADC= 50
In isosceles triangle OCD,
ODC=OCD=50
AOC=OCD+ODC -->[Ext <=sum of interior opp. <s]
=50+50
=100
ABC+BCO=180 -->[Co-interior angles]
=> 130+ BCO=180
=>BCO=50
XCO= COD+CDO -->[Ext <=sum of interior opp. <s]
={180-(50+50)} + 50
= 130
From the diagram, we see that BCO is greater than XCO
But according to the above calculations, BCO being 50 is lesser than XCO 130.
So what d'ya say??
what other method can we apply??