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?
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 $\displaystyle \angle CAD = \angle CBD$. Therefore $\displaystyle \angle CAD = y$.
From $\displaystyle \triangle DBQ$ it follows that $\displaystyle \angle BDQ = 180 - (180 - y) - x = y - x$. Then $\displaystyle \angle CDQ = x + \angle BDQ = y$.
So $\displaystyle \triangle CDA$ is isosceles etc.