Consider only "given" and "to prove" and any of the two figures.

AB is not equal to AC.

Sorry but I can't read that. (And I certainly don't see anything to do with "circles". Those look like triangles.)

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.

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, 20178 yr by studiot

Opposite interior angles of a cyclic quadrilateral are supplementary.

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, 20178 yr by studiot

Sure.

But it's hw help.... So my proof may be rejected.

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.

Fine. I shall produce the proof shortly.

Hope it's right.

Edited March 24, 20178 yr by Sriman Dutta

PEDF is not a parallelogram.

Why not? (edit)

 

You are right, it is a trapezium, CEDF is the parallelogram

 

But Sriman's proof is otherwise valid

Edited March 24, 20178 yr by studiot

Oops! Yes it isn't. It is a cyclic quadrilateral.

Oops! Yes it isn't. It is a cyclic quadrilateral.

 

#So how about the other circles question I mentioned in post#7?

Yes. I can solve it. ;-)

Edited March 25, 20178 yr by Sriman Dutta

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