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 \(MN = \frac{AB}{2}.\)

madyl Dec 20, 2019

#1**0 **

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.

madyl Dec 20, 2019

#2**-1 **

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

Guest Dec 20, 2019

#3

#4**0 **

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

#5**0 **

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

#6**+1 **

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

#8**0 **

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

#9**+1 **

I've only solved part (a)... . The address for the mathisfun solution is https://www.mathisfunforum.com/viewtopic.php?id=22435

Guest Dec 21, 2019

#10**0 **

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