Question -
If P and Q are points on circumcircle of triangle ABC such that PQ is parallel to BC prove that QA is perpendicular to simson line of P ...
My attempt -
I tried nearly 3 hours to this problem but did not get anything worth thing to conclude..... that's why I am not writing what I have got so far..... Pls give me a hint on how to start only.... and if it is possible pls also provide your diagram....thanks in advance for any help.
Let $\mathrm{SL}_p$ be the Simson lines of the point $P$. The set of all the lines and circles, except for the line $PC$, shown in Fig. 1 is the graphic description of the question posed above. We have to join $P$ and $C$ in order to facilitate the simple proof given below. Perhaps, this line is the hint that you were needing to break the stalemate. Unfortunately, we can't say this for sure, because you failed to provide us any clues. Had you attached, at least, the diagram you were staring at for three hours, someone would have noticed what you were missing in your arsenal to attack the problem.
Our aim is to prove $\measuredangle GMA = \phi = 90^0 $. To describe the proof, we need to assign letters to several points. The feet of the two perpendiculars drawn from $P$ to the sides $BC$ and $CA$ are named $F$ and $G$ respectively. The intersection of the $\mathrm{SL}_p$ and $AQ$ is called $M$. Furthermore, let $$\measuredangle MAG = \measuredangle QAC = \omega.\tag{1}$$
Because $\measuredangle QAC$ and $\measuredangle QPC$ are angles of the same segment of the circumcircle of the triangle $ABC$, $$\measuredangle QPC = \measuredangle QAC = \omega.$$
Line $PQ$ is parallel to the side $BC$ of the triangle $ABC$. Therefore, the following alternate angles are equal. $$\measuredangle BCP = \measuredangle QPC= \omega$$
Since $\measuredangle PFC = \measuredangle CGP = 90^0$, $PFCG$ is a cyclic quadrilateral. Therefore, the following angles in the same segment of the circumcircle of $PFCG$ are equal. $$\measuredangle FGP = \measuredangle FCP= \omega \tag{2}$$
Now, we know that, $$\measuredangle FGP + \measuredangle PGA + \measuredangle AGM = 180^0.\tag{3}$$
When we substitute the value of $\measuredangle FGP$ from equation(2) in equation (3), we get, $$\measuredangle AGM = 90^0 - \omega. \tag{4}$$
Now, we consider the the three angles of the triangle $AGM$, which give the following relationship. $$\measuredangle MAG + \measuredangle AGM + \measuredangle GMA = 180^0 \tag{5}$$
To complete the proof, we substitute the values of $\measuredangle MAG$ and $\measuredangle AGM$ from equation (1) and (4) respectively in equation (5). $$\measuredangle GMA = 180^0 -\measuredangle MAG - \measuredangle AGM = 180^0 - \omega - \left(90^0 - \omega\right)=90^0$$