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Two circles are externally tangent at point \(P\), as shown. Segment \(\overline{CPD}\) is parallel to common external tangent \(\overline{AB}\). Let \(M\) be on \(\overline{AB} \) so that \(\overline{MP}\) is tangent to both circles. 

(a) Prove that \(M\) is the midpoint of \(\overline{AB}\).

 

(b) Let \(Q\) be the intersection of \(\overline{AC}\) and \(\overline{BD}\). Show that \(\triangle QAB \cong \triangle PAB. \)

 

(c) Let \(N\) be the midpoint of \(\overline{CD}\). Show that \(MN = \frac{AB}{2}.\)

 Dec 20, 2019
edited by madyl  Dec 20, 2019
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BTW a hint came with the question:

 

For part (c), let \(N'\) be the intersection of lines \(QM\) and \(CD\). Show that points \(N\) and \(N'\) coincide.

 Dec 20, 2019
 #2
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I've added some more points and an extra circle.

 

If you draw the two tangents from a point outside to any circle, the tangent lengths will be equal.  So if G is the midpoint of AB, then GA = GP = GB and a circle centred on G will go through A, B and P.

 

So APB = 90 and so cos(ABP) = cos(x) = sin(PAB)

Also ABP = BPD by the parallels.

 

PB = ABcos(x) and PBcos(x) = 0.5PD  => PD = 2AB cos^2(x)

Similarly CP = 2AB sin^2(x) => CD = 2AB(cos^(x) + sin^2(x) ) = 2AB

 

When a line is tangent to a circle (at P, say) and PQ is a chord of the circle, this chord makes an angle across the other side of the circle that is equal to the angle between the tangent and the chord.  I have called this the 5th angle property of a circle (because there are 4 others) 

 

So using the property ABP = BDP and also BAP = ACP.  I know that APB = 90 (this is the 3rd angle property of a circle, same thread) so BAP + ABP = 90 => ACP + BDP = 90 as well.

So mark H = midpoint CD => AB = HD = CH => ABHC and ABDH are parallelograms.

 

By the 5th angle property of a circle ABP = BDP and BAP = ACP

So ACP + BDP = 90

 

triangles ACH, BHD and HAB are congruent (SAA) so AHB = 90 => H lies on the circle through A, B and G => GH = 0.5AB

 Dec 20, 2019
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Did you copy this from Math is Fun? I can't really understand this solution. 

 Dec 20, 2019
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Hi Madyl,

 

Looks to me like guest has put in a lot of work.

I don't know why you would ask if the answer came from Mathsisfun.  Maybe you have your reasons.

 

It is good that you have said you do not understand. It is always good to interact with answerers.

Have you spent a lot of time trying to understand?

 

My criticism is that you have not said thank you (or given a point) for the guests time and attempted explanation.

Do you think guest, or anyone else, should give you more time since you have not shown appreaciation for the time he/she has already spent?

 

 

Note:

I have not looked at the question or the answer. I do not know if the answer is correct or if it is well explained.

Melody  Dec 21, 2019
edited by Melody  Dec 21, 2019
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He simply copied and pasted this answer from the website Mathisfun. I know this because I've tried to find the answer to this question before and came across this exact post on the mathisfun forum. Technically, the guest is plagraizing so I should really thank the real writer of the guest's post I do not understand: bob bundy, an admin on Mathisfun. Basically, the guest's post was not original and he did not spend time on his/her post.

madyl  Dec 21, 2019
edited by madyl  Dec 21, 2019
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ok that is fine.

If what you say is correct then he/she was plaguizing.

You should have just accused him outright rather than asking him if he was guilty.

That does explain you attitude, thanks for that explanation.

He/she should not be rewared for plaguizing.

 

a) There is a circle propery that just says that the 2 tangential distances from an external point to a circle are equal.

Using this   AM=AP   and   BM=PM

so    AM=MB 

so if M is on AB it follows that M is the midpoint of AB

 

 

I haven't done the other 2. They would take more effort on my part.  Are you happy with my part A?

Melody  Dec 21, 2019
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Yes, I actually got it but forgot to post.

madyl  Dec 21, 2019
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Thanks for responding Madyl,

 

Do you mean you have worked out all three for yourself?

If you have then that is excellent !

 

Do you have the address for the mathsisfun solution? 

I just think if may be laid out more simply than guests copied version.

Melody  Dec 21, 2019
edited by Melody  Dec 21, 2019
edited by Melody  Dec 21, 2019
 #9
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I've only solved part (a)... sad. The address for the mathisfun solution is https://www.mathisfunforum.com/viewtopic.php?id=22435 

Guest Dec 21, 2019
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Guest, are you and  madyl the same person.

It is confusing when you do not identify yourselves properly.

 

Nevermind, thanks for the input.

I also have only done part A

That solution in mathsisfun is very confusing, I am not surprised that people cannot follow it.

 

Hopefully I will put aside some serious time and effort for this question later.  I just have to find that time and drive which might be difficult.

Someone else might answer first. 

 

CPhill is usually pretty good with these and Omi67 does a lot of geometry questions too I think.

Melody  Dec 22, 2019
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Well, thanks for your help so far anyways. laugh

For part (b) someone told me to use parallel lines... using the fact that AB||CD could be helpful...

madyl  Dec 22, 2019
edited by madyl  Dec 22, 2019
edited by madyl  Dec 22, 2019

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