Bob

]]>mathmatiKs

]]>In any circle, there are relationships between certain angles. These are called 'the angle properties of a circle'. I recently got asked to prove them all so you can look them up on:

http://www.mathisfunforum.com/viewtopic … 77#p220477

Look particularly at post #6.

I have made a simplified diagram for your question, erasing the bits that are not needed for this and marked the angles that are equal with a red dot.

Look at the chord IF. On the blue circle it makes the angle IDF and the angle IGF. So these are equal by property 2.

BAC is also equal because it is the opposite angle in the parallelogram ABDC.

Hope that helps you to understand this step in the proof.

There are three circles to 'play with' so you will be able to find lots of angles that are equal using the property. Look for a chord and a pair of angles made by the chord on the opposite side of the cirlce.

Bob

]]>angle IGF = IDF ( angles made at the circumference by the same chord IF )

=> IGE = EDF = CAB

that means?

]]>Sleeping on this problem has given a result.

I wanted to construct the circles without having to 'make' the right radius.

So I used parallelograms.

By drawing lines parallel to the sides AB, BC and CA make three parallelograms

ABDC, BCFA and CAEB.

Opposite angles of a parallelogram are equal so

ABC = DFE, BCA = FED and CAB = EDF.

So DFE is similar to ABC (letters in the correct order)

To make the circles I can use D, E and F to fix the radius.

Circle on C: Radius = CD. ( note that this circle also goes through F because AB = CD and AB = FC )

Circle on B: Radius = BE. ( for the same reason this circle also goes through D )

Circle on A: Radius = AF ( for the same reason this circle also goes through E )

So three of the points from my earlier diagram are now indentified as D, E and F .

I have labelled the remaining intersections of circles as G, H and I

My diagram shows the same similar triangles as before but using the new letters.

To show the remaining triangles are also similar I will outline just one example.

Consider triangle EIG.

In this triangle IEG = BCA (opposite angles in a parallelogram)

Draw the line IG (shown in green)

angle IGF = IDF ( angles made at the circumference by the same chord IF )

=> IGE = EDF = CAB

So EIG is similar to CBA ( two angles are the same => all three must be the same )

Hope that helps,

Bob

]]>Good luck.

Really excited to discover new things, I'm always careful.

]]>I'll try a proof tomorrow.

Bob

]]>I've just looked at the recent posts.

I have four similar triangles!

Colouring the circles helped.

See diagram.

AF = 6.30

BC=6.23

That will not be the error?Yes. I am placing the points with the mouse and 'hand shake' causes small inaccuracies.

Bob

Exactly. Are similar to the original triangle, that's what I say.

]]>I just have to prove it.]]>

I have four similar triangles!

Colouring the circles helped.

See diagram.

AF = 6.30

BC=6.23

That will not be the error?

Yes. I am placing the points with the mouse and 'hand shake' causes small inaccuracies.

Bob

]]>Thanks, I will test them for AAA. Also, I will see if it works for some other triangles. See you later.

]]>DGI FIE HFD]]>

BC=6.23

That will not be the error?]]>