Question: $\Delta ABC$ is such that $AB=AC$ and $\angle BAC=2\theta$. $D$ is a point on the same side of $BC$ as that of $A$ such that $\angle BDC=\theta.$ Is it necessarily true that the circumcircle of $\Delta BDC$ has $A$ as its centre? Prove or disprove.
This question just came out of my mind while I was thinking that, we know that if $O$ is the centre of the circle $\tau$ and $A, B$ and $C$ are three points on $\tau$, such that $C$ is on the same side of $AB$ as that of $O$, then we know that $\angle AOB=2\angle ACB$, but is the converse true? Certainly the converse is not true always. But what happens when the conditions given in the question holds true?
Proof by contradiction
Assume a circle with centre A and radius AB = AC.
Chord BC subtends angle 2theta at the centre.
As you know that any point taken on this circle will subtend half the angle it subtended at the centre, so we can clearly say that if the point D is on the circle the circumcentre of BCD will be A.
So let the contradiction be that the point D be on the outside or inside the circle and let it subtend angle theta on BC,but clearly if the point D is outside the circle it will subtend an angle less than theta and if it is inside the circle it will subtend an angle more than theta. (Can you prove it?) (And of course taking the assumption that A and D are on same side of BC)