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

Question

mycroft holmes · Answer

Call the intersection of PF and DQ as G. Since Ang GFE+ Ang GDE = 180 0 , 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 = 90 0 – 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 = 180 0 -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 = 90 0 . 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

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

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

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

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Analytical Geometry

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship