Do intersecting circles always have equal angles at the circumference?

[mattg443]

I was wondering if the common chord of two intersecting circles subtends an equal angle in both circles at the circumference (in no special cases i.e different radii circles etc)

If not, are there any special case(s) where this would work, making these two triangles similar or ABDC is a kite/special quadrilateral...

(see diagram please)

Thanks!

 

[robphy]

If you move D along its circle, the angle doesn't change.

 

[mattg443]

Yes, but is angle ADC equal to the angle subtended by the same chord in the other circle (angle ABC)

 

[LCKurtz]

If you move D along its circle, the angle doesn't change.

Yes, but is angle ADC equal to the angle subtended by the same chord in the other circle (angle ABC)

Did you try looking at the picture if you move D like robphy suggested??

 

[mattg443]

when D is in the circle, ABC=ADC

I have come up with a proof which says otherwise to robphy for when B and D are on different circles, UNLESS is it is missing a link (which is likely)

I added a centre to each of the circles and found that angle ADC is only equal to angle ABC is AO1CO2 (where O1 & 02 are the centres of the circles) is a rhombus (opposite angles equal) so that 2α=2β

Let my know the missing link I am making then for it to be true for all cases, as suggested by robphy.