[bonanova]

Triangle ABC is isosceles.

Find angle x.

Do not assume the figure is drawn to scale.

Do not use the law of sines.

Use simple geometric reasoning, like

• 
  
• Isosceles triangles have two equal angles and sides.
  
• The angles of a triangle sum to 180 degrees.
  
• Things like that.
  

Have fun.

You can add a line or two if you think it will help.

[Guest]

Just to stop the pain in my head, I plugged this geometry into AutoCAD, and the answer is definitely 20 degrees. But for the life of me, I can't find the answer without using sine/cosine or some other method.

but that is just it

you cant find it without using those.

edit:

might i mention thank you for giving me the best/worst migraine of my life

Edited September 13, 2008 by solemnraven

[bonanova]

Tossing a couple Excedrin to SR.

Top to bottom:

Angle at C: 20^o

Angles at E: 150^o-x, x, 30^o

Angles at D: x+10^o, 130^o-x, 40^o

Angles at crossing point: 50^o, 130^o, 130^o, 50^o

Angles at A; 10^o, 70^o

Angles at B: 20^o, 60^o

You can use triangle congruency [sAS] and [ASA] and the like.

Just keep it geometrical - no sine and cosine formulas.

And you can read the OP spoiler.

[Guest]

I surrender... please give the solution...

[bonanova]

A vertical line thru point C that bisects the big triangle.

A line from point A that intersects BC at a point the same height as point D.

You get two pairs of congruent triangles and the solution.

[bonanova]

Draw the red lines to create points L, M, N:

[1] Angle CAB = Angle CBA [triangle ABC is isosceles] = (180-20)/2 = 80

[2] Angle ACL [20/2] = Angle CAL [80-60-10] = 10. So triangle ACL is isosceles.

[3] AL = CL

[4] Angle LAM = Angle ECL = 10

[5] Angle ALM = Angle CLE [opposite angles of intersecting line segements]

[6] Triangles ALM and CEL have two angles[4,5] and included side[3] equal thus are congruent [Mirror Image].

[7] EL = LM [corresponding sides of congruent triangles.]

[8] CM = CL+LM and AE = AL+LE [inspection]

[9] CM = AE [3,7,8]

[10] Angle ACN (20) = Angle CAN [equal to CBD(20) by symmetry]. So triangle ACN is isosceles.

[11] AN = CN

[12] Angle EAN [80-60-10] = Angle MCN [20/2].

[13] Triangles AEN and CMN have two sides[9,11] and included angle[12] equal thus are congruent [Mirror Image].

[14] EN = MN [corresponding sides of congruent triangles.]

Angle BAN = Angle ABD(60) [symmetry] so triangle ABM is equilateral.

DN is parallel to AB [AN is symmetrical to BD] so Triangle DMN is equlateral.

[15] Angle DNE = Angle ABE = 80

[16] DM = DN = MN

DN = EN [14, 16] so Triangle DEN is isosceles.

[17] Angle EDN = Angle DEN = (180 - DNE)/2 = (180-80[16])/2 = 50

[18] Angle AEB = 180 - (Angle EAB + Angle ABE) = 180 - (70[given] + 80[15]) = 30

So finally,

x = Angle DEN - Angle AEB [inspection] = 20 [17, 18].

[k-man]

That's a really cool puzzle. Keep them coming

[Guest]

The best puzzle here . 5 Star from me , I was going around in circles

[Guest]

i think we all were....almost go tme an SR on a office beating rampage

bring em on again

[Guest]

It's really cool! Way to go! Bona!

[andromeda]

I loved it as well (I almost put here two s's ) , love geometry!

Give us some three dimensional ones!

[Guest]

I was going thru the solution minutely and found this

[4] Angle LAM = Angle ECL = 10

Angle LAM needs to be proven to be 10.

This proof is simple so I guess you missed a few steps ... anyways, this was a great puzzle

[Guest]

Great puzzle and a much greater solution... Nice one bona!!!

Waiting for #2!!!

[Guest]

Solved ...

why are those complications guyz its so easy

Edited September 18, 2008 by Mohamed Adelestien

[Guest]

x=60

[Guest]

Solved ...

why are those complications guyz its so easy

The diagram and angles you have are wrong, look again

[bonanova]

Angle LAM needs to be proven to be 10.

Missed it

Angle MAB = 60 [equilateral => equiangular => 60]

Angle EAB = 70 [given]

Angle LAM = Angle EAN = Angle EAB - Angle MAB = 70 - 60 = 10.