muskan Posted March 8, 2017 Share Posted March 8, 2017 Consider only "given" and "to prove" and any of the two figures. AB is not equal to AC. Link to comment Share on other sites More sharing options...
Country Boy Posted March 8, 2017 Share Posted March 8, 2017 Sorry but I can't read that. (And I certainly don't see anything to do with "circles". Those look like triangles.) Link to comment Share on other sites More sharing options...
muskan Posted March 9, 2017 Author Share Posted March 9, 2017 ABC is a triangle. P, Q and R are mid- points of AB, AC and BC respectively. AD is the altitude from A to BC. Prove that PRDQ is cyclic. Link to comment Share on other sites More sharing options...
muskan Posted March 9, 2017 Author Share Posted March 9, 2017 Link to comment Share on other sites More sharing options...
studiot Posted March 11, 2017 Share Posted March 11, 2017 (edited) I really think you should practise drawing half way decent diagrams. Yours are , frankly, not good enough to work from. Here is a better one. Note that I have labelled all 14 angles, even though I don't yet know if I am going to use them. Now first question what is your strategy for proving that PQDR is cyclic? What property of the angles of a cyclic quad do you know? If you do not know the answer to this ask as it is the key to the proof. Once this is answered you can assemble the necessary information. There are 5 triangles, 3 quadrilaterals and 4 points where some of the angles add up to 180. These will give you lots of very simple equations between some of the angles. But not enough. You should get used to using all the information provided in a question. This analysis has not yet used the fact that P, Q and R are midpoints. What do you know about lines joining midpoints of triangle sides? Applying this will yield enough further equations to reduce the number of unknown angles since many can be shown to be equal with this extra information. I don't see that the negative mark in post#3 is either productive or warranted so I have added a +1. Edited March 11, 2017 by studiot 1 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 23, 2017 Share Posted March 23, 2017 Opposite interior angles of a cyclic quadrilateral are supplementary. Link to comment Share on other sites More sharing options...
studiot Posted March 23, 2017 Share Posted March 23, 2017 (edited) Opposite interior angles of a cyclic quadrilateral are supplementary. That's correct, Sriman. So can you complete the proof? Can you also complete this second one? http://www.scienceforums.net/topic/103844-circles/#entry975863 Edited March 23, 2017 by studiot Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 23, 2017 Share Posted March 23, 2017 Sure. But it's hw help.... So my proof may be rejected. Link to comment Share on other sites More sharing options...
studiot Posted March 23, 2017 Share Posted March 23, 2017 Sure. But it's hw help.... So my proof may be rejected. I don't think anyone will mind now, the question was so long ago that the OP cannot claim credit at school for it. There have been 163 views so we are addressing others who may like to know. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 23, 2017 Share Posted March 23, 2017 Fine. I shall produce the proof shortly. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 24, 2017 Share Posted March 24, 2017 (edited) Hope it's right. Edited March 24, 2017 by Sriman Dutta Link to comment Share on other sites More sharing options...
muskan Posted March 24, 2017 Author Share Posted March 24, 2017 PEDF is not a parallelogram. Link to comment Share on other sites More sharing options...
studiot Posted March 24, 2017 Share Posted March 24, 2017 (edited) Why not? (edit) You are right, it is a trapezium, CEDF is the parallelogram But Sriman's proof is otherwise valid Edited March 24, 2017 by studiot Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 24, 2017 Share Posted March 24, 2017 Oops! Yes it isn't. It is a cyclic quadrilateral. Link to comment Share on other sites More sharing options...
studiot Posted March 24, 2017 Share Posted March 24, 2017 Oops! Yes it isn't. It is a cyclic quadrilateral. #So how about the other circles question I mentioned in post#7? Link to comment Share on other sites More sharing options...
Sriman Dutta Posted March 25, 2017 Share Posted March 25, 2017 (edited) Yes. I can solve it. ;-) Edited March 25, 2017 by Sriman Dutta Link to comment Share on other sites More sharing options...
Minato Posted May 29, 2017 Share Posted May 29, 2017 You can also solve the same problem with little shorter method if You make use of circumcircle covering a right angle triangle which will be trianglr ADC and Q and mid point of AC making AQ=QC=CD hence Angle QDC=angle QCD for isosceles triangle QDC and angle RPQ=angle QCD now we will talk all in angle so ill stopt typing angle before all sets RPQ + RDQ =RPQ +RDA+ADQ =QCD+QDA+ADR =CDQ+QDA+ADR =180 as all 3 angle lies on a line hence we can say PQDR is cyclic Link to comment Share on other sites More sharing options...
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