Tricky geometric proof

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Wildcard
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Tricky geometric proof

Postby Wildcard » Wed May 20, 2015 8:18 am UTC

Given:

Code: Select all

A̅B̅ ≅ D̅B̅ ≅ C̅B̅
D̅E̅ ⟂ C̅B̅
A̅F̅ ⟂ C̅B̅
E̅G̅ ⟂ D̅B̅
∠ABD ≅ ∠DBC

Proof diagram.png

Prove:

Code: Select all

A̅F̅ ≅ E̅G̅


I think this one is tricky, personally. There is an alternate method of solution which involves
Spoiler:
trigonometry and use of the unit circle,
but there is in fact a standard solution without going beyond geometry and conventional methods of geometric proofs.

(I thought of putting this one in the "Mathematics" forum but this forum heading says, "For good logic/math puzzles.")

EDIT: Apparently the givens aren't showing up correctly on some displays. Thank you to Sandor for clarifying it for everyone. I like my diagrams uncolored :) so I'm restating the puzzle in words below.

Given:
AB = DB = CB
DE is perpendicular to CB
AF is perpendicular to CB
EG is perpendicular to DB
angle ABD is congruent to angle DBC

Prove:
AF = EG
Last edited by Wildcard on Wed May 20, 2015 9:20 pm UTC, edited 1 time in total.
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Sandor
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Re: Tricky geometric proof

Postby Sandor » Wed May 20, 2015 12:15 pm UTC

The same diagram, but now with Paint goodness (apologies to the colour blind)
Painted proof diagram.png
Painted diagram
Painted proof diagram.png (36.07 KiB) Viewed 5570 times

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SPACKlick
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Re: Tricky geometric proof

Postby SPACKlick » Wed May 20, 2015 2:58 pm UTC

Your given has some display issues, could you reframe it in words?

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jaap
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Re: Tricky geometric proof

Postby jaap » Wed May 20, 2015 3:11 pm UTC

SPACKlick wrote:Your given has some display issues, could you reframe it in words?

Sandor's picture has all the info in. I couldn't see some of the characters in the OP either, but it was fairly easy to deduce they must have been the symbol for perpendicular, but that has now all been captured in Sandor's picture.

sfwc
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Re: Tricky geometric proof

Postby sfwc » Wed May 20, 2015 3:48 pm UTC

Spoiler:
Let K be the intersection of AC and BD. Note that AC and BD meet at right angles because ABD = DBC and BA = BC. We have ACF = KCB = 90 - KBC = 90 - DBE = BDE and CFA = 90 = DEB, so that the triangles ACF and BDE are similar. Let K' be the intersection of GE and BD. Then ABC = GBE and ACB = KCB = 90 - KBC = 90 - K'BE = K'EB = GEB, so that ABC and GBE are similar.

Now AF/AC = BE/BD = BE/BC = GE/AC, so that AF = GE = EG.

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SPACKlick
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Re: Tricky geometric proof

Postby SPACKlick » Wed May 20, 2015 3:52 pm UTC

Spoiler:
BE = AB Cos(a)
DE = AB Sin(a)
EC = AB-BE
EDC = arctan(AB-ABcos(a)/AB Sin(a)) = a/2
DE = (AB-ABcos(a))*(cos(a/2)/sin(a/2)) = AB*sin(a)
Define H as the midpoint of EG, necessarily where it crosses BD
EH = AB*sin(a)*cos(a)

EG = 2(AB*sin(a)*cos(a))=AB sin(2a)
AF = AB sin(2a)

I don't have notes but here's my proof. It's different from sfwc's

Or if I read the question which looks for a geometric proof...dammit. I leapt for trig.

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Re: Tricky geometric proof

Postby Wildcard » Wed May 20, 2015 9:43 pm UTC

Wow, nice answers! Sorry about the formatting issues; the symbols show up as perfect geometric notation on a Mac. I've edited the OP now but left the older notation in place also.

@ SPACKlick: I didn't check every step of your proof, but I don't blame you for jumping to trig! I'm 99% sure your answer is equivalent to my trig answer. In actual fact, trig is where I came up with the puzzle:
Spoiler:
If you regard the diagram as being drawn on the unit circle (with radius AB), then it is a general diagram of the fact that sin 2x = 2 sin x cos x.
After I made the diagram, knowing that the answer was true (that AF = EG), I set about trying to prove it with geometry. I finally got it after it bounced around in my head for a couple days. (See below spoiler.)

@ sfwc: Wow, nice! I never thought of that approach! My geometric approach was very different:
Spoiler:
Define P as the midpoint of AB and define Q as the midpoint of BD. Define circle P to have radius AP (so it goes through A and B) and define circle Q to have radius BQ (so it goes through B and D). Notice that circles Q and P are congruent, as they have congruent diameters and therefore congruent radii. (AB and BD are the diameters of circles P and Q respectively.)

Now notice that circle P passes through point F. This is because angle AFB is 90 degrees and intercepts a 180 degree arc of circle Q, demonstrating that AFB must be an inscribed angle.

Notice also that circle Q passes through points E and G, by identical reasoning to the previous line.

(I may not be stating this part of the proof in the clearest way possible. The geometric fact is that an inscribed angle always intercepts an arc with double the measurement of the angle, and any angle is an inscribed angle which intercepts an arc with double its measurement. There's probably a cleaner way to state it but that's the essence of the lemma and it has been proven in basic geometry books many times over.)

Continuing, since angle ABF is congruent to angle GBE, and since these angles are inscribed angles in each of two congruent circles, the chords intercepted by these angles must also be congruent. Therefore AF = GE. QED.
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RCT Bob
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Re: Tricky geometric proof

Postby RCT Bob » Thu Aug 27, 2015 8:43 am UTC

I have a different proof

Spoiler:
For ease, call angle ABD "a", like in Sandors picture. I'm calling lengths AB and BD "r", as the radius of the circle that A, D, and C are on with midpoint B. And I'm calling the intersection of lines BD and EG " H".
Now then, because angle DBE is also a and AF is perpendicular to BF, I can write
[math]AF = r sin (2a)[/math]

I can also prove EH = HG

[math]HG = EH = BH tan(a)[/math]

Next up, because DE is perpendicular to BE:

[math]BE = r cos(a)[/math]

Next, because EH = HG:

[math]EG = 2 EH =2 BE sin(a) = 2 r cos(a) sin(a) = r sin (2a) = AF[/math]

By double-angle formula for the sine

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Re: Tricky geometric proof

Postby Wildcard » Sat Apr 16, 2016 3:21 am UTC

Nice work, RCT Bob.

I guess by now you've read the spoilers....

Good you filled in the gaps, although I think it's more straightforward to say:
Spoiler:
Designate a unit distance such that BA = BD = BC = 1 unit.

Define theta as the measure of angle ABD (and likewise the measure of angle DBC).

Note that DE = sine theta, and BE = cosine theta.

Define point H as the intersection of GE and BD.

By similar triangles, note that HE = sine theta cosine theta.

Also note that GH = HE, again by similar triangles (AAAS).

Therefore EG = 2 sine theta cosine theta.

Note that AF = sine 2 theta.

Now by basic trigonometry we see that AF = EG.


Last bumped by Wildcard on Sat Apr 16, 2016 3:21 am UTC.
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