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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.

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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.


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

Edited by solemnraven
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Tossing a couple Excedrin to SR. ;)

Top to bottom:

Angle at C: 20o

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

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

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

Angles at A; 10o, 70o

Angles at B: 20o, 60o

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.

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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].

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