Geometery problem Isoceles triangle?

This is an old, well-known and beautiful problem, here is a solution, not using trigonometry:

http://farm5.static.flickr.com/4067/4579289396_25a...

Triangle BCE is 80-50-50 (all angle measures below are in degrees), hence

|BC| = |BE|. Take a point E' on AB such that angle(BCE' ) = 60 and let F is the intersection point of BD and CE'.

Triangles BCF and DE'F are equilateral, so |BF| = |BC| = |BE|, i.e. triangle BEF is 20-80-80, triangle CEF is 10-30-140, hence

angle(EFE' ) = 40 = angle(EE'F) since angle(AE'D) = 80 and angle(DE'F) = 60

Consider the quadrilateral DE'EF - it is a kite - a union of 2 isosceles triangles: DE'F and EE'F. DE is its axis of symmetry (on the picture both symmetric parts are cyan and lime-green) and halves the angle BDE'. The latter is 60, hence angle(BDE) = 30.

See another nice problem about 20-80-80 triangle:

http://answers.yahoo.com/question/index;_ylt=AuifL...

P.S. Excellent link, Mathematishan!

Reading the article reminded me a late friend of mine (we studied Mathematics together in the University long ago), who had collected several problems about triangles 20-80-80, 40-70-70, 80-50-50, 100-40-40 and 140-20-20. There is some charm in such problems, due to the fact that trigonometric functions of π/9 and some its multiples are not expressible in square roots, so inventive synthetic solutions are preferable.

Easiest way to start is to draw your triangle.

Let me call the intersection point where BD and CE meet point F.

So we know that BAC is 20, so the other angles ABC and ACB are each (180-20)/2 = 80 degrees

We also know that DBC is 60, so ABD = 80 - 60 = 20

and that ECB is 50, so ACE = 80 - 50 = 30

FBC is 60, and FCB is 50, so BFC is (180 - 50 - 60) = 70

EFB is 110, because the angle at point F along CE must be 180 by definition, and we know the 70 degrees already, so the other angle must be 110

Similarly, EFD is 70 and DFC is 110.

If EFD is 70, then the other two angles DEF and FDE total 110

BEC is 180 - 80 - 50 = 50

BDC is 180 - 60 - 80 = 40

All sorts of angles, but not quite what we're looking for, so we must use the sine rule (check Wikipedia for the definition).

sin60 / DC = sin40 / BC = sin80 / BD

... so DC = sin60 * BC / sin40

sin50 / BC = sin50 / BE = sin80 / CE

... so CE = sin80 * BC / sin50

sinCDE / CE = sinCED / CD

... sinCDE / sinCED = CE / CD = [ sin80 * BC / sin50 ] / [ sin60 * BC / sin40 ]

= [ sin80 * BC / sin50 ] * [ sin40 / sin60 * BC ]

= (BC * sin80 * sin40) / (BC * sin50 * sin60)

= (sin80 * sin40) / (sin50 * sin60)

= 0.954...

So the ratio of sinCDE to sinCED is (sin80 * sin40) / (sin50 * sin60)

or that sinCDE = 0.954... * sinCED

and you know that DEF and FDE total 110

... not quite sure how to get the very last step, but maybe you can?

i assume you're speaking with regard to the degree of the angles. Assuming you're working in ranges: provided that an Isosceles triangle has a minimum of two equivalent angles and that your base angles are 3x - 11 and 2x + 11 3x - 11 = 2x + 11 or x - 11 = 11 or x = 22 that could advise that the backside angles are fifty 5 ranges. because of the fact the sum of the angles of any triangle equivalent one hundred eighty ranges, then the the rest attitude is 70 ranges. From this we get here equation provided that the pronounced length of the the rest attitude is 2y: 2y = 70 y = 35

After drawing the triangle that maverick has explained I agree with his angles how ever i end up getting EDB=55°