Post Number:#2 by joydip » Fri Jan 13, 2017 1:10 pm
A generalization of the problem :
Let $P$ be the intersection point of sides $AD$ and $BC$ of a convex qualrilateral $ABCD$. Suppose that $I_1$ and $I_2$ are the incenters of $\triangle PAB $ & $\triangle PDC$, respectively. Let $O$ be the circumcenter of $\triangle PAB$, and $H $ the orthocenter of $\triangle PDC$. Show that the circumcircles of $\triangle AI_1B$ and $\triangle DHC$ intersects at an angle $\theta$ if and only if the circumcircles of $\triangle AOB $ and $\triangle DI_2C $ intersects at an angle $\theta$.
Proof:
Lemma : Let $\alpha$ & $ \beta$ be two circles meeting at point $A , B$. If $C \in \alpha$ & $ D \in \beta$, then the angle between $\alpha$ & $ \beta$ is equal to the angle between $(ADC)$ & $(BDC)$.
proof :Let an inversion with center $D$ and an arbitary radius take $A,B,C$ to $A',B',C'$ respectively .Then the angle between $A'C'$ & $B'C'$ is equal to the angle between $A'B'$ & $(A'B'C')$ by alternate segment theorem ,completing the proof .
Let $(AI_1B),(DI_2C),(DHC)$ meet $BC$ at $M,N,S$ respectively .Let $(AI_2B) \cap (DHC)={ U,V }$.
Let $(BVC) $ & $(AVD)$ meet again at $X$.
$\angle DXC= \angle VAD +\angle VBC=\angle VAD + \angle MAV=\angle MAD=\angle DNC $ So, $X \in (DI_2C)$
$\angle AXB= \angle VXB + \angle AXV= \angle VCB + \angle VDA=\angle SDV + \angle VDA= \angle SDA=2\angle APB=\angle AOB $. So, $X \in (AOB)$
Let $(BUC) $ & $(AUD)$ meet again at $Y$.Then Similerly we get $(DI_2C) \cap (AOB) ={X,Y}$
So, by our lemma angle between $(AUV)$ & $(DUV)$ , angle between $(AUD)$ & $(AVD)$,and angle between $(AXY)$ & $(DXY)$ are same, completing the main proof.
(the original problem can be solved more easily, only by considering $X $ & $Y$)