Problem 6...................... Just to reach minimum character limit)

Call the intersection of PF and DQ as G. Since Ang GFE+ Ang GDE = 180⁰, GFED is a cyclic quad and hence G also lies on the incircle.

 

Now join FD. Let Ang FED = b. Then  it easily follows that Ang FQD = Ang FDP = 90⁰ – b.

 

That means FQPD is a cyclic quadrilateral (angles made by the chord FD in the same segment are equal).

 

Further the segment FG makes an angle of 2b at the incentre and since B is the point of concurrency of tangents BF and BD, it follows that Ang FBD = 180⁰-2b = 2 Ang FQG. This implies that B is the centre of the circle through FQPG. So we get that BQ = BP. Also note Ang PFQ = 90⁰.

 

That means QP is a diameter i.e. it passes through B, so that Q, B,P are collinear and B is the mid-point of PQ as required to be proved