1. ## Circle Geometry 3

Hi

DA produced and CB produced meet at Q. If angle CDB = angle DQC = x, prove that CD = ca. (Hint: Let angle CBD = y)

Can anyone please give me a hint on how to do this question?

2. Originally Posted by xwrathbringerx
Hi

DA produced and CB produced meet at Q. If angle CDB = angle DQC = x, prove that CD = ca. (Hint: Let angle CBD = y)

Can anyone please give me a hint on how to do this question?
I don't usually do geometry but since my first post was a circle geometry one and since it's the anniversary of my first post I will do this one.

There's a theorem that tells you that $\angle CAD = \angle CBD$. Therefore $\angle CAD = y$.

From $\triangle DBQ$ it follows that $\angle BDQ = 180 - (180 - y) - x = y - x$. Then $\angle CDQ = x + \angle BDQ = y$.

So $\triangle CDA$ is isosceles etc.