# Do this geometries. It R Harder

• Continuing the discussion from Do this maths. It R Hard.:

Using only Algebra and Elementary Geometry (No trig, calc, etc.), finc the value of `x` (Angle DEA) in the triangle below.

Nota Bene: to prevent cheating the triangle is not drawn to scale.

• It R Harder

<you expected me to work here?

• Something tells me I should know this, but it's been more than a decade now since I did this stuff, and I'd probably go to calculus if I needed an answer. Set A=(0,0), B=(1,0), derive the other points from there.

• I've worked out a load of the angles, and got it down to a set of 4 simultaneous equations. I'm not entirely sure that's enough to solve it but if it isn't then either the solution is unfindable or I'm missing something

• I'm missing something

i'm not sure if it's required, but the solution i used required constructing additional triangles by thge careful adding of points and then using triangle symmetry to prove that angle X was the same as a different angle x' that i knew the angle of.

• Adding all the numbers that can be trivially calculated:

Next step: a set of simultaneous equations involving w, x, y, z.

And now to wait for people to explain how I worked those angles out, despite it being maths taught at the age of six…

• I don't think I've done this wrong:

[spoiler]
ABC: 80 + 80 + C = 180
C = 20
80+60+D = 180
D = 40

ABE = 70 + 80 + E = 180
E = 30

ACE = 10 + 20 + E = 180
E = 150
(150 - x) + 30 = 180
150 - x = 150
x = 0
[/spoiler]

• You can get most of the angles, but nothing involving the DE line because you don't know any angles involving that line. Angles ACB, ADB, AEB, AEC, BDC can be trivially derived. But to get further than that you probably need to construct helper triangles that are similar to each other.

You can definitely get there through calculus. Dx = n, Dy = ntan(80°), Dy = (n-1)tan(60°), is solvable. Same for Ex and Ey. And once you have the coordinates getting all the angles is trivial.

• Next step: a set of simultaneous equations involving w, x, y, z.

That's where I got to. No parallel lines for fun with angle symmetry, so I pretty much lost interest

• I dun goofed. (150 - x) + x + 30.

• Hmm....I figured out x, but I kind of made a leap that I'm not happy with and couldn't find a way around it. But I didn't try new triangles...which sounds more likely than that I forgot some angle relationship.

• I can fill in the missing angles, but then I have to guess based on that.
Can't figure out how to make a proper equation out of this.

• Next step: a set of simultaneous equations involving w, x, y, z.

``````w = 150 - x
w = 160 - z

x = 130 - y
x = 150 - w

y = 130 - x
y = 140 - z

z = 160 - w
z = 140 - y
``````

OK, the equations are sorted… and I don't think you can solve that set.

• So tempted to start an argument about something right in that that I can convincingly claim is wrong...

• Belgium ed up my math

• Good news: I was able to get to a 4x4 system of equations as well -- it's not terribly hard if you're at all conversant in triangle properties.

Bad news: that system is underspecified.

• These equations, yes?

``````w = 150 - x
w = 160 - z

x = 130 - y
x = 150 - w

y = 130 - x
y = 140 - z

z = 160 - w
z = 140 - y
``````

All I can prove from that is `x = x`

• The trouble is that that's actually only 4 equations (w = 150 - x and x = 150 - w are equivalent), which with 4 variables isn't enough to get any results. You can go round in circles and get a few tautologies though

• Where did I Belgium up in my system of equations? Wolfram also agrees that that set cannot be solved.

• I would like to name the spot in the middle of the triangle "STEVE", but that would be confusing. It can just be "F instead.

The sum of all three angles in a triangle is 180, and so is the sum of two angles on a straight line. That's all we need here, so let's fill in some angles.

EAB=70, EBA=80 so AEB=30.
If AEB is 30 then AEC = 150.
ACB=20, CBD=20 so BDC = 140.
FAB=70 and FBA=60 so AFB=50.
If AFB is 50 then DFA = 130.
On the other side of that line, DFE=AFB=50 and EFB=DFA=130

Now we have most of the triangle DFE.
DEF=x, DFE=50 and FDE=130-x

Since CDB =140 and EDB=130-x, CDE=x+10

Here's the part where I used a time machine to give myself the answer. Triangle CDB is the same as CED. That pins angle CED to 140, which makes DEF=10, CDE=20 and FDE=120.

So, x=10. All I need to do now is remember why those two triangles are congruent, and then tell myself.

(Remembering how spoiler tags work would be helpful too. But that can wait.)

• I got about 3-4 angles in before I realized it's not a simple a " remember a triangle is 180 degrees and fill in the blanks"-type problem. Then I said I don't care anymore and gave up.

• Found where I belgium ed up. Accidentally flipped a sign.

• Triangle CDB is the same as CED.

This is the bit I'm not following; I'm not sure how you'd derive that

• This is the bit I'm not following; I'm not sure how you'd derive that

He's simply asserting it is without proof by then borrowing a joke from Bill and Ted's Excellent Adventure.

• My solution:

[spoiler]
Note that by simple addition CAB = CBA = 80 degrees
Plot new point bisecting AB label R
ARC = BRC = 90 degrees (because of CBA = CAB
Draw new point D' by mirroring across line CR
Draw new Point E' by mirroring across Line CR
Draw new point O on intersection of lines E'B and EA
AOB is similar triangle to E'OE
Line E'E is parallel to Line AB
Angle AOB = 40 therefore Angle E'OE = 40
Line E'E and line CR intersect at right angles.
therefore Angle E'EO is 70 Degrees (confirmed by simmilar triangles)
Create point Q on intersection of line DE and D'E'
Create Poitn P on intersection of DB and AD'
APB = 60 degrees so D'PD = 60 degrees
Because of construction E'QE is similar triangle to APB
Therefore E'QE = 60 degrees, and since APB is equilateral and similar to E'QE angle E'EQ is 60 degrees
E'EA is 70 degrees from above E'EQ is 60 degrees from above so QEA is 10 degrees.
Line DE intersects point Q therefore DEA= 10 degrees
[/spoiler]

• This is the bit I'm not following; I'm not sure how you'd derive that

When you find out, let me know. It made perfect sense at the time, but I can't prove it.

Er... That is to say... "I have discovered a truly marvellous proof of this, which Discourse is too narrow to contain."

• At first, I thought you ass-pulled figures, but they do indeed check out; @accalia later confirmed that with her own solution, which is purely geometric (I tried algebraic; it can't be done that way)

• i'm not claiming my asnwer is the shortes t solution, but it does hold

• Flaw in your solution right here:

Line E'E and line CD intersect at right angles.

Line CD lies along line CB.
Angle CBA is 80°.
Line E'E is parallel to line AB.
∴ angle CEE' is 80°

• whoops. typo's there. that should have been line CR not CD.

• My solution:

1. Open http://web.geogebra.org/app/
2. Build the triangle with the data I have
3. Use the angle tool.

Work smart people!

[spoiler][/spoiler]
geogebra.ggb (8.7 KB)

• Because of construction E'QE is similar triangle to APB

That's quite a leap. What are you basing this claim on?

• hmm.... i should have written my notes down, at home now and i did the original diagram on work computer. let me see......

AOB/E'OE are similar with vertex touching.

D and D' are symmetrical and used to construct both points P and Q which forms a doubled pair of similar triangles inside AOB/E'OE

which because the doubled pair are all similar and subtracted symmetrically from a third pair of similar triangles the newly constructed E'QE and APB must be similar.

• AOB/E'OE are similar with vertex touching.

Good.

D and D' are symmetrical and used to construct both points P and Q

True.

which forms a doubled pair of similar triangles inside AOB/E'OE

which because the doubled pair are all similar and subtracted symmetrically from a third pair of similar triangles the newly constructed E'QE and APB must be similar.

No. This presumes that DE and D'E' are parallel to AD' and BD, respectively. There is not enough information to support this logical leap.

• /me looks that up......

FRACK

i KNOW i have the right answer and i know can prove it..... I did it this morning for fracks sake.

/me gets out pencil and paper

i'm not going to sleep until i have this.

• A wise fox once told me...

turn off the computer and go to bed before i call [your significant other] and have [them] take the computer away.

• my bed time isn't for another hour and i'll have this by then.

• Good, because 4,000 miles is a long way to travel just to take away your computer

• Here's what I drew before looking at the thread. It adds to @RaceProUK's diagram by having the new letters in different places, more colors, and an underspecified system of equations that I hope I have solved correctly so far. To be honest, I'm not sure where to go from here, though I've not looked in detail at the proposed solutions...

(The colors indicate logical inference order: a color needs (& only needs) the information in the colors that are to it's left in Paint's palette. )

• Never mind. I screwed up.

• ICHIBAN! I HAVE IT!

so i was hitting dead end after dead end trying to complete my proof. I know i could have (altough: [spoiler]my answer of X=10 was wrong[/spoiler]) So i struck out in an entirely new proof and finally got it.

[spoiler]

1. Draw line || AB from D Label intersection with CB Point P
2. By #1 Triangle DCP ~ Triangle ACB
3. By #2 Angle CPD = 80 degrees
4. By #2 Angle ADP = 100 degrees
5. By #2 Angle BDP = 60 degrees
6. Draw Line PA, label Intersection with DB Point Q
7. Angle APD = Angle BDP
8. Angle DQP = 60 degrees
9. Triangle DQP is equilateral
10. Triangle ABQ is equilateral
11. Angle ACP = 20 Degrees and Angle CAP = 20 degrees so Triangle ACP is isocoles
12. Draw line CQ, which bisects Angle ACB (AP == DB so intersection Q is placed such that CQ bisects ACB)
13. Triangle ACQ ~ Triangle CAE
14. Line PC - Line CE = Line PA - Line AQ = Line PE = Line FQ
15. by #14 Line PQ = Line PD so Line PE = Line PD
16. with two equal side Triangle DPE Must Be Isosceles
17. Angle DEP = 30 + x = (180-80)/2 = 50
18. By #17 X = 20
19. @‍accalia was wrong about X = 10. whoopsies.
[/spoiler]

i would have had that sooner but..... well it took a lot of paper to make sure i knew the proof and that when i cut our the dead ends i still had all the information required to actually complete the proof (you will want to follow along the proof with paper to see everything adds up)

• I agree with your answer. However, I arrived at it differently.

[spoiler]I cheated. I drew it in Inkscape and measure the angle. [/spoiler]

• Looks good but:

15. by #149

FTFY

• Where do you get #8 from?

• Easiest way is to extend BD out to D' such that BCD' is isosceles, then you know the angles of regular trapezoid BEDD', which gives you x.

• #8 is by....

missed a step. bugger.

Angle BDP = 60 so by reflection (AB and DP are parallel and Triangle ABC is isoscoles) so Angle APD = 60.
AQP are colinear by the construction of Q
DQB are colinear by the consstruction of Q
thus Angle DPQ = Angle QDP = 60
with two angles = 60 Triangle QPD must be equilateral and so Angle DQP = 60

insert that between 7 and 8

• Work smart people!

OK then.

20°

hmm.... i should have worked harder to make sure you didn't get sensible results.

• Can we get extra credit for selfies with sex toys?