Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

Two circles are externally tangent at point P. Segment CPD is parallel to common external tangent AB. Prove that the distance between the midpoints of AB and CD, is AB/2.

only without the thin drawn in lines. I couldn't find a better photo.

Girls can be just as good as boys at math. We just need to get the same encouragement.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi championmathgirl

Here's my diagram:

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

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

Bob

Sorry this is a bit 'round the houses'. I've done this problem before and I don't think I did it exactly the same so it might be worth researching that thread.

EDIT Could only find this

http://www.mathisfunforum.com/viewtopic.php?id=21335

which is not the version I was looking for.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

Thanks, but i don't really understand how you got AB = HD = CH or how you got ACP + BDP = 90.

Girls can be just as good as boys at math. We just need to get the same encouragement.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

I have shown that CD = 2AB and H is the midpoint, so CH = half CD = AB and HD similarly.

As AB is parallel to CH and to HD this makes ABHC and ABDH parallelograms.

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)

I have proved these properties in another post. I have been asked to help with a lot of geometry recently, including the same question more than once. I'm not complaining; I like to be helpful; but I was wondering whether it would be helpful to members if I created a post consisting of a geometry contents list with links to the relevant posts. What do you think of this idea?

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.

If you think it will be helpful to create the 'contents' list I'll research the right thread as the first entry for you.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Dmaharjan****Member**- Registered: 2015-08-12
- Posts: 2

Hello Bob,

What do you mean by "If you draw the two tangents from a point outside to any circle, the tangent lengths will be equal. " Could you explain a little more on it.

Thanks.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi Dmaharjan

A picture is worth a thousand words:

T is a point outside a circle, centre C. Two tangents, TA and TB are drawn to the circle.

Theorem: AT = BT

Proof: Consider the triangles ATC and BTC.

TC is common. TAC = TBC = 90. CA = CB = radius.

So the triangles are congruent (SSA A=90), and so AT = BT.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**championmathgirl****Member**- Registered: 2015-06-01
- Posts: 20

Sure, contents list would be helpful!

Quick last question, if ACH, BHD and HAB are congruent (SAA) how is it that AHB = 90?

Girls can be just as good as boys at math. We just need to get the same encouragement.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

If they're congruent, then they have the same angles, and we know that two of the angles add up to 90.

Bob

Contents list at

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Dmaharjan****Member**- Registered: 2015-08-12
- Posts: 2

Hello Bob,

Thanks that helped.

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi Bob,

While trying to solve the puzzle I found this property (maybe you know it), that the red lines from AB's midpoint G are tangents at P & H of the red circle, whose centre O lies on the same line as centres E & F of the two green circles.

That then means that GH = GP = AG = GB, from which GH = AB/2. Simple. Just one tiny hurdle, though: I can't find any way of including this property in a proof.

Also, PJ = JH, which again would give us GH = GP = AG = GB, from which GH = AB/2. But I don't know how to prove that PJ = JH.

Have you got any clues on these two issues, or is the solution route going to be too long to be worth pursuing? I'd hoped to find a relatively quick answer to the puzzle, but it has eluded me.

Thanks,

phro

*Last edited by phrontister (2017-02-26 23:35:51)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi phro,

I hadn't spotted that. Once I've done the problem my way, I can easily show the properties you want, but that doesn't help if you want to develop a fresh solution. I've had a longer and more thorough search for an earlier version of this problem and found nothing. So I think I must have dreamed that I've seen it before.

I feel there must be an easier way to do this (it's only CompuHigh after all ) so I'm going to start again with a fresh diagram. Wish me luck.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Good luck, Bob!

Yes, I thought there should be an easier way, too...as there was with this one, with ΔPEB being 1/3rd of ΔPCA.

Here's another diagram, with another property revealed by Geogebra: ie, CL = LP = HM. But again, I don't know how to use that info.

This time I've marked some of the equal-length lines.

*Last edited by phrontister (2017-02-26 23:34:24)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

Hopefully what follows is a useful way to do this problem. It's still a little long but loads of useful properties drop out during the proof. One of the problems when the diagram is complicated and accurate is it becomes very easy to assume something that needs to be proved. I think I've ironed out any such bugs but please let me know if you spot any **.

In this diagram AB is parallel to CD as given. E and F are the circle centres. G is the midpoint of AB.

GA = GP = GB by the equal length tangents theorem. So a circle centred on G will go through A, B and P.

It will also cut the line CD again whilst the original circles are unequal in size. Let H be the point where the circle cuts CD. Continue HG back across the circle to point L.

Let angle GAH be x and angle GBH be y.

As H is subtended by a diameter AHB = 90, and as BGH and AGH are isosceles, GHB = y and GHA = x, and x + y = 90.

Let BF cut CD at J.

As ABF is 90 and AB is parallel to CD, BJC = 90, and so PJ = JD.

Let AE (extended in my diagram) cut CD at K.

Then, in the same way, AKD = 90 And CK = KP.

So ABJK is a rectangle.

CD = CK + KP + PJ + JD = 2(KP + PJ) = 2KJ = 2AB.

I have just spotted a **. I have assumed that DB produced and CA produced meet at L. I shall have to prove that.

Everything falls into place once I've done this so there will be a (hopefully short) intermission while I get this step sorted.

OK. I think I've got it.

Angle LGB = AGH (vertically opposite) and triangle GBL is isosceles so LBG = x.

At point B we have angles of x, y, and x so JBD = y and BDJ = x. So DB extended goes through L. Similarly CA extended so L is the point where DB and CA cross.

Lots of results follow:

(1) triangle LAB is similar to LCD and half its size.

(2)LBHA is a rectangle.

(3)ABDH has a pair of parallels and opposite angles equal so it is a parallelogram.

(3)Similarly, ABHC is a parallelogram.

(4)So it follows that H is the midpoint of CD.

(5)GH is half AB.

Bob

Bob

*Last edited by bob bundy (2015-11-23 23:02:03)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi Bob;

That extension to L has done the trick, I reckon!

Here's my revised proof, with most of it pinched from yours. I've streamlined it (and the image) as much as I could, but maybe you can see something else...or a blunder!

=========================================================================================

In this diagram AB is parallel to CD as given. E and F are the circle centres. G is the midpoint of AB.

GA = GP = GB by the equal length tangents theorem. So a circle centred on G will go through A, B and P.

It will also cut the line CD again whilst the original circles are unequal in size. Let H be the point where the circle cuts CD.

Let angle GAH be x and angle GBH be y.

As H is subtended by a diameter, AHB = 90°.

Because A, H & B are on G's circumference and AHB = 90°, AHBL is a rectangle with L on the circle so that HGL is a diameter.

AHB = 90°, so x + y = 90°. Therefore LBA = x and BAL = y.

LB extends to D because HBD = 90° and LBA = BDH = x (AB being parallel to CD); similarly, LA extends to C (HAC = 90° and BAL = HCA = y). Therefore BAL = DHB = HCA = y, and LBA = AHC = BDH = x.

Triangles LBA, BDH and AHC are congruent (similar angles, LA = BH and LB = AH). Therefore their hypotenuses are equal to each other (CH = HD = AB), proving that H is CD's midpoint.

We already know that GH, AG and GB are radii of circle G, and therefore GH (ie, the distance between the midpoints of AB and CD) = AB/2.

*Last edited by phrontister (2017-02-26 23:32:27)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi phro,

Nice one. Just one point:

AHBL is a rectangle with L also located on G's circumference.

You need to say AHBL is a rectangle with L on the circle so that HGL is a diameter. Only then will HBL and HAL (remember him?) be 90.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi Bob;

Yes - good point. I've changed my post.

Only then will...HAL (remember him?) be 90.

HAL? How could I forget him?! If his incorrect interpretation of Daniel 9:27 and his associated prediction were, in fact, correct, I wouldn't be writing this now! I knew he had the wrong angle/slant on that: he's 86, not 90! Had his birthday on Monday.

*Last edited by phrontister (2015-11-24 11:13:15)*

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi phro,

We are several million miles apart. This is the HAL that I was referring to:

But hey! Never mind. I spent some educational moments reading about quotes from the bible and misinterpretations of those. Conspiracy theorists may be able to find a connection, given the film title and subject matter.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi Bob,

I never saw "2001: A Space Odyssey", but I do know "Space Oddity" by DB, who, according to our images, nearly meets HAL. That is eerily significant, because (according to Wikipedia) DB features in a book published by none other than HAL LEONARD CORPORATION...and maybe they even met!!

Conspiracy theorists' favourite hobby is to create seemingly plausible connections where none exist.

*Last edited by phrontister (2015-11-25 02:23:43)*

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

Many people found the film incomprehensible and I would too, except I am a keen Arthur C Clarke fan, and I had read the story "The Sentinel", on which the film is based. I can recommend this story and it makes sense. According to Wiki it was published in a collection of short stories under the same name.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**evene****Member**- Registered: 2015-10-18
- Posts: 272

Referring to phrontister's diagram, couldn't you just instead prove that

is isosceles with ?*Last edited by evene (2015-11-25 05:34:45)*

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi evene,

...couldn't you just instead prove that

is isosceles with ?

GP = HG as they are radii of circle G. This means that ΔGPH is isosceles (unless PGH is a straight line).

From post #14:

(a) GP: "So a circle centred on G will go through A, B and P." Therefore GP is a radius of circle G.

(b) HG: "It will also cut the line CD again whilst the original circles are unequal in size. Let H be the point where the circle cuts CD." Therefore line HG is a radius of circle G.

Up to that point H hasn't been proven to be CD's midpoint, but that is done in the second half of post #14.

If H is set as CD's midpoint first - ie, instead of H being the intersection of CD and circle G as per (b) above - it's probably not all that straightforward to then prove that HG = GP.

*Last edited by phrontister (2015-11-26 01:21:05)*

Offline

**evene****Member**- Registered: 2015-10-18
- Posts: 272

I don't know if you already thought of this, but can we also extend BD and AC to meet at point Q above AB, and then somehow use similar triangles to prove that what needs to be proven?

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

I don't understand what you mean there, evene. The upward extensions of those two lines already meet above AB at point L in my diagram.

Point L features prominently in relation to the congruent triangles used to prove that H is CD's midpoint.

*Last edited by phrontister (2015-11-27 10:46:15)*

Offline

**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,594

Hi all;

Here's another approach:

The circles (centres E & F) meet at P; CD is parallel to common external tangent AB; G is AB's midpoint; H is *nominally* CD's midpoint.

Proof that H is *actually* CD's midpoint:

(i) ABML is a rectangle (*tangent perpendicular to radius* theorem for points A & B; AB is parallel to CD; AE extends to L), and is bisected by GJ.

(ii) ΔPGH is isosceles (GH = GP), J is base PH's midpoint, and PJ = JH = **y**.

(iii) CL = LP = **x** (*perpendicular from centre bisects chord* theorem).

(iv) HM = LJ - PJ = **x** (J is LM's midpoint; PJ = JH).

(v) MD = PM = **x** + 2**y** (*perpendicular from centre bisects chord* theorem).

(vi) CH = HD = 2**x** + 2**y**, and so H is CD's midpoint.

GP = GA = GB (*equal length tangents* theorem), and, from point (ii) in the proof above, GH = GP.

∴ GH (ie, the distance between the midpoints of AB and CD) = (GA + GB)/2 = AB/2

*Edit 2/9/16: I changed point labels (text and image) to bring them into line with previous posts' labels. The points affected are L (was K), J (was L) & M (was J). Sorry...should have been a wake-up to that.*

*Last edited by phrontister (2017-02-27 11:24:20)*

Offline

**basketballstar123****Member**- Registered: 2016-08-10
- Posts: 11

phrontister wrote:

Hi Bob;

LB extends to D because HBD = 90° and LBA = BDH = x (AB being parallel to CD); similarly, LA extends to C (HAC = 90° and BAL = HCA = y). Therefore BAL = DHB = HCA = y, and LBA = AHC = BDH = x.

Triangles LBA, BDH and AHC are congruent (similar angles, LA = BH and LB = AH). Therefore their hypotenuses are equal to each other (CH = HD = AB), proving that H is CD's midpoint.

We already know that GH, AG and GB are radii of circle G, and therefore GH (ie, the distance between the midpoints of AB and CD) = AB/2.

https://onedrive.live.com/download?resid=C20C46B976D069EE!3932&authkey=!AHpqkk_oTZr-SGg&v=3&ithint=photo%2cjpg

... But how to you know HBD=90????

Offline