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

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

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1 Answers

mycroft holmes
272 Points
4 years ago
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

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