Another hardest easy geometry problem?

Update:

fizixx, nice try but a claim does not a fact make

Update 2:

James D, you want an "elegant solution"? You're kidding, right?

Update 3:

Well, Fred, I'll leave this one open for a while then, "for further investigation". Can't wait to see an "elegant solution".

Update 4:

falzoon, it seems to me that you've shown that ΔADE = ΔBFG, but why does ∠ADE = ∠BFG = 60 necessarily?

Update 5:

Josh Swanson, I like your approach because it suggests a general means of coming up with other examples of this kind of problem. I think any good "elegant solution" should point the direction to generalization, rather than being "elegant but only for a very particular case".

Update 6:

Fred, you'd nominate falzoon's answer as BA? Wow, let me have another look at his answer.

I'll be closing this out soon, and then I'm going to post another question on this subject of "hardest easy geometry" problems of this type.

Update 7:

Falzoon

Update 8:

Falzoon

Update 9:

(Boy, is Y!A really BUGGY!)

Falzoon & Fred, can either one of you explain why x = 60 necessarily? We know that ΔADE = ΔBFG and ∠ADE = ∠BFG, but so? Why is x = 60, "therefore"?

Update 10:

Falzoon, sure, I'll give this plenty more time if folks want it open.

Update 11:

I'm going to extend this question so that I can more carefully analyze some of your answers.

12 Answers

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  • Duke
    Lv 7
    7 years ago
    Favorite Answer

    Happy New Year to all in Y!A Mathematical Forum!

    Back here from my Christmas vacation I found this excellent problem and couldn't resist the temptation to try myself on it. Imagine a regular 18-gon P₁P₂P₃...P₁₈ , inscribed in a circle with center O. Its vertices divide the circumference of the circumcircle into 18 equal arcs, containing 20° each, as shown here:

    http://farm8.staticflickr.com/7304/11706199485_a31...

    Now draw the chords P₁P₇ and P₅P₁₁ (blue) - they have equal length because the subtended arcs are (7 - 1)*20° = (11 - 5)*20°. Let Q is their common point. The symmetry considerations show that the diameter through Q is their axis of symmetry, so the lines P₁P₇, P₅P₁₁ and OP₆ are concurrent.

    Draw the chord P₁P₆ (dark green) now:

    ∠P₆P₁P₇ = ½ arc_measure(arc P₆P₇) = 10°

    The angle between chords P₁P₆ and P₅P₁₁ is

    ½ (arc_measure(arc P₅P₆) + arc_measure(arc P₁₁P₁₈P₁)) = ½ (20° + 8*20°)

    so they are perpendicular. The latter fact, along with

    ∠P₁P₆P₅ = ∠P₁P₆O = 40°

    shows that P₅ and Q are symmetric with respect to P₁P₆, or, say it differently, ∆P₆P₅R (light-gray) is congruent to ∆P₆QR (gray; R is the common point of lines P₁P₆ and OP₅). But then ∠OP₅Q = ½ arc_measure(arc P₁₁P₁₄) = 30° and

    ∠P₅RP₆ = ∠P₆RQ = 60°.

    The green triangle OP₁P₆ (100°-40°-40°) is your one with the given angles 10°, 20°, 30° and 80°.

    * * * * * *

    EDIT: To everybody - see what similar problem I dug out about the same 40°-40°-100° triangle ABC (equal angles at A and B): M is an internal point, such that ∠MAB = 10°, ∠MBA = 20°. Connect M with C and find the angles at M by ELEMENTARY reasoning.

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  • Fred
    Lv 7
    7 years ago

    There are 15 angles in the diagram (not counting those that "cross over," compounding two other angles).

    4 of them are given; one more is the unknown, X; 6 of the remaining 10 are pretty easy to compute from triangles, and straight angles, where in each case, the sum must be 180º.

    fizixx has done this; I have checked his values for those 6 angles and concur with them.

    He has also filled in the remaining 4 angles, including X, and I differ with him on those. I will label those 4 angles as follows.

    Where "X" is labelled, label the next angle clockwise from it, "W." (The next angle clockwise from that is 60º.)

    In the uppermost, "little" triangle, the upper angle is 40º. One of the lower two angles is X; label the other one, Y. Counterclockwise from Y, label it, Z. (Counterclockwise from Z is 50º.)

    We can now write the following 4 equations in the 4 unknown angles:

    (1) W + X = 120º

    (2) X + Y = 140º

    (3) Y + Z = 130º

    (4) W + Z = 110º

    Unfortunately this is a redundant set, and contains only 3 independent equations

    [e.g., (1) - (2) + (3) = (4)].

    After more ugliness than I can stand to chronicle here, using the Law of Sines, the Law of Cosines, the equality of co-functions of co-angles, and the double-angle formulas, I arrive at

    tanW = sin40ºsin60º/(2sin20ºcos40º - (1 - cos40º)sin60º) = 1.7320508076... = √3

    W = 60º

    X = 120º - W = 60º

    Y = 140º - X = 80º

    Z = 110º - W = 50º

    PS: I would be astonished if there isn't a much neater way to get this.

    And yes, Scythian, I guarantee there IS a more elegant solution.

    I can also just about guarantee that I can't come up with it in any kind of reasonable time.

    These assertions are intuition-based, not mathematics-based.

    Add.Det.:

    Well, Scy., I wasn't talking days there. More like months.

    I actually rather like JamesD's solution. That was the other approach I had in mind -- analytic geometry -- when I started down this path. But he's already carried it out.

    But, good problem! It really has that, "there's bound to be a shorter way" quality. And it has led to this highly improbable trig identity:

    sin40º = 2(2sin20ºcos40º - (1 - cos40º)sin60º)

    EDIT:

    Define

    s = sin20º; c = cos20º

    S = sin40º; C = cos40º

    Then

    S = sin40º = 2sc

    1 - cos40º = 2s²

    c = cos20º = cos(40º-20º) = Cc + Ss

    sin60º = sin(20º+40º) = sC + cS, and

    2sin20ºcos40º - (1 - cos40º)sin60º

    = 2sC - 2s²(sC + cS)

    = 2s(C - s²C - scS)

    = 2s(c²C - scS)

    = 2sc(cC - sS)

    = S cos60º = ½S

    Thus

    2(2sin20ºcos40º - (1 - cos40º)sin60º) = S = sin40º

    verifying that

    tanW = sin40ºsin60º/(2sin20ºcos40º - (1 - cos40º)sin60º) = 2sin60º = √3

    . . just killing time while not coming up with anything elegant.

    EDIT2:

    OK, Scythian, you can close it now.

    IMHO, falzoon has the long-awaited elegant proof.

    And Josh gets the red ribbon.

    EDIT3:

    Why must X=60º? Corresponding parts of congruent triangles!

    . . . ∠X = ∠ADE corresponds to ∠BFG, which is 60º by construction.

    To me, the beauty of falzoon's solution is that he spotted a symmetry the rest of us overlooked (or at least, I did!) -- that the overall triangle is isosceles, because

    [the upper angle] = [the leftmost angle] = 40º.

    From there, it's a simple step to extend that symmetry by drawing CF and FG, and the rest falls right into your lap.

    BTW, I totally sympathize with your principle that one aspect of elegance of a solution, is the scope of its generality.

    And Josh's solution has that.

    EDIT4:

    Hold it! Josh has a point!

    The essential problem seems to be that ∠ABE=10º is what determines the location of E, while ∠BFG=60º determines the location of G, its putative symmetric partner.

    So I think you'd have to draw a line from A, at 10º inside AB, and show that it must cross BC at G. This can be done, but it's a little more work.

    2013-12-29 Sun 21:05:39 +0000

    EDIT5:

    All right. Duke has lifted the rest of the veil!

    I've gone over his steps, and can verify all of them; some by different means than he uses -- it's all good.

    Your ∠X = his ∠P₆RQ = 60º

    2014-01-02 Thu 18:40 +0000

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  • 7 years ago

    It seems to me this is really about the quadrilateral in the image linked. In the image's notation, from the angle addition formula for sin and the law of sines,

    sin(110) cot(W) - cos(110)

    = sin(110-W)/sin(W)

    = p/s

    = p/r r/q q/s

    = (sin 10/sin 60)(sin 80/sin 30)(sin 50/sin 20)

    (Every angle in the image can be found just by adding up angles to 180, which also shows that X=120-W, so finding W is sufficient.) Since cot is monotonic, this has a unique solution. Hence we just need to check W=60 satisfies this equation. At W=60, multiply both sides by the denominator of the RHS to get

    sin(110-60) sin(30) sin(20)

    = sin(10) sin(80) sin(50),

    hence proving X=60 reduces to showing sin(20) sin(30) = sin(10) sin(80). Using standard identities,

    sin(20) sin(30)

    = 2sin(10) cos(10) sin(30)

    = 2sin(10) sin(80) 1/2

    = sin(10) sin(80),

    as needed. This seems reasonably elegant.

    Edit: @Scythian/Falzoon: I don't see how we know ΔADE = ΔBFG. It's clear to me this is equivalent to x=60. Put another way, if you replace "Then draw |FG| at 60º to |BF|" with "Then draw |FG| at 70º to |BF|," what step breaks? As far as I can tell you haven't actually argued your situation is symmetrical, you've just asserted it.

    Edit: @Fred: I'm pretty sure "but it's a little more work" is not true. I don't see how to patch the hole without some involved reasoning. We seem to require one non-obvious relation, and many suffice (eg. I computed W instead of X), but I don't see it coming from just the simple observation that the triangle is isosceles. Perhaps Falzoon will have more to say.

    Edit: @Quadrillaterator: Thanks for the careful write-up. I checked your proof and found no flaws, typographical or otherwise. It took quite a while, and my final diagram is a real mess; I can see why you didn't include a picture. I agree, I'm sure it can be made cleaner and briefer, but I at least won't take the time. It's interesting what low-level tools suffice to solve this problem. I don't know how this method would solve the general problem. Perhaps I'm biased, but I greatly prefer my own solution, since it's so straightforward to verify, solves the general problem, and is still quite basic.

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  • nerey
    Lv 4
    4 years ago

    Easy Geometry

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  • 4 years ago

    For the best answers, search on this site https://smarturl.im/aDV8h

    Let f=AF so that AE=2f, and let Δ = area of triangle ABC. Then area of triangle AEF = (1) AEF = f²Δ/(12*16) But we can write AEF in another way: (2) AEF = ½Δ(AG/AM)(2f/16) + ½Δ(AG/AM)(f/12) = ½Δ(AG/AM)f(5/24) so that (3) AG/AM = f/20 Now we find two ways to express AEG: (4) AEG = AEF(EG/EF) = (EG/EF)f²Δ/(12*16) (5) AEG = (½Δ)(AG/AM)(2f/16) Equating (4) to (5) and plugging in (3) yields (6) EG/EF = 3/5 so that (7) EG/GF = 3/2 Explanation: The important thing going on here is that if you have an arbitrary ΔPQR such that P' is on PQ and R' is on QR then the area of ΔPQR = PQR * (P'Q/PQ) * (R'Q/RQ). This is easy to see because the area of a triangle is half base times height, and each time we do one of those ratios, we are treating that side (the one we're getting the ratios from) as the base. By the way, there is a second way to go on this problem, which is sort of a quick and dirty approach... Since you know that the ratio is constant, you can pick your own favorite AE or AF. So make it something convenient, like put E=C! Then AF=8. So why is this useful for us? The answer is that if you have the ratio of any two of the six sides and cevians of a triangle, you can easily get the rest. What's a cevian, anyway? It's just a segment that starts at a triangle's vertex and goes to the opposite side. Now there are two theorems that it's worth memorizing (and they're easy to memorize). The first theorem (Ceva's Theorem) is that if you've got an internal point P in a triangle so that the three cevians through P cut the sides in ratios of m:n, p:q, and r:s, then mpr=nqs. Pretty nifty, huh? The ratios have to be taken in the same order. So the idea is that you go around the triangle labelling each of the 6 cut side segments with its length or ratio. Then multiply every other one and you get the same product in both cases. For example, if you have a triangle with two medians intersecting, they cut their sides in ratios of 1:1 and 1:1. Therefore, the other side will be cut in a ratio of 1:1, too. The problem with this example though is that it doesn't make the ratio ordering very clear, so let's use the problem at hand. With C=E, let H be the intersection of BG with AC. Now write down, next to the triangle, CM=1, MB=1 (you don't know their actual lengths, but you only need the ratios for this), FB=4, AF=8. Then our theorem tells us that CM*FB*AH = MB*AF*HC, or 4AH = 8HC. In other words, HC/AH = 1/2. Now, the second theorem (an easier to remember relative of Menalaus' Theorem) is a little more involved: If you've got a cevian included between two sides whose ratios you know, then the ratio of the upper part of the cevian to the lower part of the cevian is equal to the sum of the ratios of the upper part of the two sides to the lower part of the two sides. For example, where you have the three medians of a triangle as the cevians, the ratio of the upper part of any of those cevians (medians) to the lower part is 1/1 + 1/1 = 2. In other words, medians cut each other in a 2 to 1 ratio. Applying the 2nd theorem to your problem, it says EG/GF = CG/GF = CH/AH + CM/MB = 1/2 + 1/1 = 3/2. Well, it's quick and dirty if you already know the theorems. In that case, it's about two lines. :) One more 'by the way': this problem also falls in the category of mass point geometry. The third reference below has some examples that illustrating your particular type of problem.

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  • 7 years ago

    Diagram at: http://s205.photobucket.com/user/falzoon/media/Ang...

    First, it's obvious that <BAC = 40º and <BDC = 60º.

    As in the diagram, draw |CF| at 20º to |CB|.

    Then draw |FG| at 60º to |BF|.

    Then, <FCD must be 60º and therefore, <CFD = 60º.

    We now have the following -

    ΔABC is isosceles and includes equilateral ΔFCD, centered, so that

    ΔFGC is congruent to ΔDEC, and ΔADE is congruent to ΔBFG.

    Therefore, x = <BFG = 60º.

    EDIT: Scythian, if you're closing out soon, I'll have to leave you with Fred's comments,

    as I've got to go to work. Won't be able to come back to this for about 12 hours.

    EDIT: I do see the problem now. I've assumed just a little too much from the fact

    that triangles FGC and DEC have only 1 angle and 1 side in common. They need

    at least 1 more fact to be congruent. I think I can do it, with elegancy suffering,

    so I'll get back a bit later, as now I have a birthday dinner to attend.

    EDIT: Let's start again, with a new diagram -

    http://s205.photobucket.com/user/falzoon/media/Ang...

    Draw the 3 construction lines -

    |CF| at 20º to |BC|, |FG| at 60º to |AB| (facing towards point B) and |AG| at 10º to |AB|.

    From ΔABC, calculate <BAC = 40º. ΔABC is thus isosceles, with |AC| = |BC|.

    <BCD given = 80º. Therefore, <DCF = 60º.

    Because of the symmetrical 20º angles, |DC| = |FC|.

    Now from ΔBCD, calculate <BDC = 60º.

    Then from ΔCDF calculate <DFC = 60º.

    Thus, ΔCDF is equilateral, and by symmetry, |AD| = |BF|.

    From ΔBCE, calculate <BEC = 50º.

    From ΔBFG, calculate <BGF = 80º.

    From the angles at point A, calculate <CAG = 40º - 10º = 30º. Then, <AGC = 50º.

    Δ's BCE & ACG are congruent because of 3 equal angles and (as above), |BC| = |AC|.

    Therefore, |CE| = |CG|. This now proves that Δ's CDE and CFG are congruent, as

    each has two sides the same, with an included same angle.

    Therefore, |DE| = |FG|, being the third side of each triangle.

    Also, |AE| = |BG|, because |CE| = |CG| and |AC| = |BC|, as shown above.

    We now have Δ's ADE and BFG congruent, as each have the same 3 sides with

    one equal angle.

    Now I can say (with just a little trepidation) that x = <BFG = 60º.

    I originally thought the most elegant way to go would be by using only geometry, as

    I've seen this done before on a few similar problems. Hopefully, I've got the job done

    correctly this time around. It's a bit longer than I'd hoped for, but it only requires

    construction lines, angles and simple geometric principles. A nice problem.

    EDIT: I do believe you are correct Quadrillerator. I do not know that <GAB = 10º.

    That firmly takes me out of the ballgame and I haven't got any time to try and fix

    that one. Too little time and not enough rigour. Thanks for waiting, Scythian.

    EDIT: Al P has thankfully persisted to where I think I can save a little grace.

    Instead of carelessly calling <GAB = 10º, I can call it y, as it is unknown at this time.

    From my 2nd diagram, <CFG = 60º. Therefore, in ΔAFG, <AGF must = 60-y, and so,

    consequently, <AGC = 40+y.

    I have previously shown that the figure is symmetrical with two sets of congruent

    triangles either side of a central equilateral triangle, so all corresponding sides and

    angles must be equal. Thus, <BEC = <AGC, that is, 50 = 40 + y, so y = 10º.

    Then x immediately follows. Hoping that is rigorous enough. Am I back in now?

    EDIT: On the flip side, I'm particularly taken with Josh's answer,

    which can be reduced fairly simply. We have, initially:

    sin(110)cot(W) - cos(110) = sin(10)sin(80)sin(50) / [sin(60)sin(30)sin(20)]

    On the RHS, replace

    sin(80) with cos(10), sin(60) with √3/2, sin(30) with 1/2 & sin(20) with 2sin(10)cos(10).

    On the LHS, replace sin(110) with cos(20) and cos(110) with -sin(20).

    After substitutions and cancellations, we get:

    cos(20)cot(W) + sin(20) = 2sin(50)/√3

    But, sin(50) = sin(30 + 20) = sin(30)cos(20) + sin(20) cos(30) = cos(20)/2+√3sin(20)/2

    Therefore, 2sin(50)/√3 = cos(20)/√3 + sin(20)

    Subtract sin(20) from each side giving cos(20)cot(W) = cos(20)/√3

    Cancel cos(20) to give finally, cot(W) = 1/√3 or tan(W) = √3, so W = 60º.

    EDIT: Once again, Quadrillerator, you have confounded me, as I didn't go through

    all the previous steps before I put forward the update, where hopefully, I would have

    realised that |CE| is not yet proved to be equal to |CG|. I should have started afresh.

    You are correct that congruency has not yet been established for Δ's BCE and AGC.

    Back out of the ballpark again, I shall humbly sit on the sideline watching proceedings.

    Your shorter proof appears sound, but after my rash moves and now, mind overload, I

    don't feel qualified to judge its merit. I'm just glad I'm not in Scythian's shoes.

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  • nle
    Lv 7
    7 years ago

    Using my notation, I have the following equations

    60 + a +X =180

    a +b +70 = 180 ( equation 2 )

    X+Y = 140 ( Y is the angle next to b)

    50 +b +Y =180

    Solving 4 equations above then X= 10 +b

    Now let BC =1 unit

    So : 1/sin(60) = AB /sin (80 ) ==> AB =1.137158043

    Also : BD/sin(100) = 1/sin(50) ==> BD = 1.285575219

    Now BD/ sin(a+60) = AB/sin(b) ( equation 4 )

    Now solve equation 2 and equation 4 then a=60 and b=50

    X = 10 +b =60 degrees

    See the link: ( for reference )

    http://s794.photobucket.com/user/smith123chart/med...

    ------------------------------------------

    <GAB is in fact 10 deg as obvious from the symmetry

    Also refer to falzoon's 2nd diagram

    FG is || with DC and DE is || with FC ( I verified by analytic geometry)

    but how to prove it geometrically ?

    Once you prove it geometrically, the rest is easy.

    And also if <BFG = 60 then X =60 deg from symmetry.

    So far only falzoon's solution is purely geometry.

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  • fizixx
    Lv 7
    7 years ago

    This is what I claim.

    *** Ok....I know, should have been more specific in that regard, but instead of going back and changing my response I will leave it stand as is......no prob. I dislike responders that post a final answer and then go back and make changes based on criticisms. However my numbers are fact, they do check, I just did not include a 'solution set'. Oh well.

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  • JamesD
    Lv 4
    7 years ago

    The problem seems to be fully defined; my geometry sucks so I shamefully bashed the **** of out it.

    If base length is 1, equations of lines from base vertices are y = (tan 40)x, y = (tan 30)x, y = (-tan 80)(x - 1) and y = (tan 80)(x - 1).

    Then points of intersections are ( [tan 80]/[tan 40 + tan 80] , [(tan 40)(tan 80)/[tan 40 + tan 80] )

    and ( [(tan 80]/[tan 40 + tan 80] , [(tan 30)(tan 80)/[tan 80 - tan 30] )

    The gradient of the line joining the points of intersection is found to be tan (-20).

    So x = (10 + 30) - (-20) = 60 degrees.

    Please reveal a more elegant solution.

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  • 7 years ago

    This was a hard problem to find an simple solution to, and I suspect that it could be even easier, but ..., no trig, no ratios, no law of sines, no law of cosines in what follows. Only congruent triangles and the occasional plus and minus. I'll start off using the diagram that Falzoon has provided at http://s205.photobucket.com/user/falzoon... but there will be some additional construction to do on it.

    As did Falzoon, construct F on AB to give equilateral ΔCDF. I don't use his FG or point G, but there will be a few others momentarily. Now define three quantities for convenience:

    AD=BF=s (for smallest),

    AC=BC=b (for biggest (or base, take your pick)) since ΔABC is isosceles.

    CD=DF=CF=e (for equilateral)

    Now extend CD to J so that ∠JAD is 40. Then ∠CJA=80=∠JAC so that JC=b.

    The goal of this proof is to show that ΔADJ = ΔADE (that will be my notation for congruent triangles), which will be done when I show that AJ=AE, whereupon ∠ADE's value is immediate.

    Pick K on AC such that BK is the angle bisector of ∠ABC, and extend BK to L such that BE extended is the perpendicular bisector of AL. Thus, AB=BL=2s+e, and AE=EL. It looks like DEL is a straight line, but we can't assume it.

    Let H be the intersection of BK with CD. Then isosceles ΔCBH gives us ∠BCH=∠BHC=80, BH=b, ∠BHD=∠BCK=100, ΔBCK=ΔBHD, and BD=BK=s+e. Furthermore, by marking off M on BH such that ΔCKM is equilateral, we have that ΔBCM=ΔCBF so that BM=e, and CM=MK=CK=s, and HD=s, too, from the congruency of ΔBCK=ΔBHD that we just saw.

    We have that CH=JA=e-s (CH since CD=e and HD=s, and JA since ΔBCH=ΔACJ), and JD=b-e (ie. CJ-CD). In addition, KL=s (since ΔABL is isosceles, BL=2s+e and BK=s+e). Thus, since ΔADJ and ΔLKE are both 40-60-80 triangles with the side opposite the 80 degrees being the same (AD=KL=s), ΔADJ=ΔLKE so that AJ=EL=AE=e-s. This means that EK=JD=b-e, and that ΔDAE=ΔDAJ=ΔKLE by SAS, which makes ∠ADE = 60 (the quantity we are asked to prove) and oh yea, D-E-L is a line with DE=JD=b-e. QED.

    Update:

    The essence of this proof is that we're rotating the broken path BCKLE 20° CCW about B onto path BHDAJ.

    Because there are so many line segments running around in this problem, I'll summarize them:

    AD=BF=CM=CK=MK=KL=DH=DK=s

    FC=CD=DF=BM=e

    AC=BC=BH=CJ=b

    AB=BL=2s+e

    BD=BK=s+e

    JD=EK=DE=b-e

    AK=b-s

    JA=CH=EL=AE=e-s

    HK=e+s-b

    CE=s+b-e

    In addition, there are some interesting relationships between b, e, and s:

    1. e(2s+e) = b(s+e)

    2. s(s+2e) = be

    For 1, compare sides opposite angles 40, 80 on similar ΔADJ vs. ΔBDC.

    For 2, compare sides opposite angles 20, 40 on similar ΔCBF vs. ΔBAK

    Other interesting relationships (such as (e+s)(e-s)=sb, e(s+e)=s(s+e+b), e³-s³=3es²) derive from the first two.

    If someone can tell me where there is a freely usable online editor whereby I can make a diagram, I'd appreciate it. I can't download software onto my machine.

    Falzoon: How do you know that the third segment, starting off from A at 10° impinges at G? ie. How do you know ∠GAB = 10°?

    Josh: Glad you checked over my proof, I've looked over yours, too, and it's straight and to the point and handles the general situation. I think both solutions have their place, and context dictates the preferred one - in this case, what did Scythian have in mind? I interpreted "easy geometry" as meaning simplest geometric principles.

    Falzoon: You have shown that ΔCDF is equilateral, hence centered over the altitude to BA (since ΔCDF is isosceles, this altitude is also a median). Since ∠BCF=∠ACD=20, you concluded that ΔBCF=ΔACD (eg. by SAS). However, I don't see that you've established that ΔBCE (a 30~100~50 triangle) is congruent to ΔACG (a 40-y~100~40+y triangle). Perhaps an alternate view is to say that if, instead, angle x HAD been given as 60, but neither ∠ABE nor ∠BAG had been given, how would one determine ∠ABE? In that case, the symmetricity would be given, but to determine the requested angle, length relationships need come into play (whether by simple or analytic geometry).

    Update: Here's a shorter proof, starting from Falzoon's original diagram (warning: though the letters are the same, I suggest not using the diagram generated by the earlier proof as it's easy to get confused):

    We already have by construction that ΔCDF is equilateral. Now mark off K on EC such that AD=DK. This means ∠AKD=∠CAD=40 so that ∠CDK=20=∠ACD. Thus, AD=DK=KC.

    Now place M such that CM=KC and ∠MCB=40. Then ΔBCM=ΔCBF by SAS so that BM=CF=FD, and also ∠BMC=120. Since ∠KCM=100-∠MCB=60 and since CM=CK, ΔCKM is equilateral so ∠CMK=60. Since ∠KCM and ∠CMK are supplementary, BMK is a straight line of length FD+AD, which is also the length of BD.

    This means that ΔDBK is isosceles with ∠DAK=20 making BE its angle bisector so that BE is the perpendicular bisector of base DK. Thus DE=EK, so that ΔBDE=ΔBCK. Now ∠CBK=20, so ∠BKC=60, so ∠BKA=120=∠BDE. Finally, ∠x=180-∠BDE=60. QED

    Update: Yet a shorter proof:

    Starting with just isosceles ΔABC with ∠C=100, mark off E on AC so that ∠EBC=30. Pick P such that BE is the perpendicular bisector of CP, and let N, R be the intersection of CP with BE and BA, respectively. This makes ΔCPB equilateral with CB=CP=CA. Right ΔCNE=ΔPNE implies CE=PE and ∠ECP=∠CPE=40. Let D be the intersection of AB with EP. Since ∠CAB=40 and ∠AEP=80, this makes ∠ADE=60.

    If we could only say that ∠ACD=20, then we would be done with ∠X=∠ADE. To that end, since ΔCEP=ΔCRA (by ASA), CE=PE=AR=CR so that AE=PR. Since ∠CPE=40=∠CAB and ∠PRD=80=∠AED, ΔPRD=ΔAED (by ASA), so that RD=ED. Thus, ΔCRD=ΔCED (by SSS) so that ∠ACD=∠RCD=½∠ACP=20. QED

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