Pure geometry question?

Let E so that B is on [AE] and CAE isosceles in A. Then BE = DA.

Let's call this length L.

Let O be the circumcenter of CBE. The angles of COB are (20,20,140), COE (10,160,10) and BOE = (60,60,60). Hence L is the circumradius of CBE.

Let F inside ABC so that ABF is congruent to CBO and G inside ABC so that CBG is congruent to CBO.

All angles can be computed and DAF, AFG, FGC are all congruent to COE. It follows that DAFGC all lie on a circle. Hence

angle(CD,CA) = angle(CA,CF) = angle(CF,CG) = 1/2 *angle(CA,CG) = (1/2)*20° = 10°.

Edit. Maybe you should ask Duke. He is the expert on pi/9 geometry.

Edit 2. You are right. There is much simpler.

Let E so that EBC is equilateral. Then angle(BE,BD ) = 100° - 60° = 40°.

Since CA = BD and BE = CB, BED is congruent to CBA.

It follows that ECD is isosceles in E with top angle 160°.

Hence angle(CE,CD) = 10°, so

angle(CD,CA) = angle(CE,CA) - angle(CE,CD) = 20° - 10° = 10°.

Though that one is simpler, I much prefer the first proof because the extra constructions are much more revealing.

Thanks, Mathematishan, for the invitation to answer, I like such problems very much.

Follow the link below to see my solution:

http://farm5.static.flickr.com/4144/5056997500_da8...

P.S. Not the simplest possible solution, but with 3 isosceles triangles 20°-80°-80°, 80°-50°-50° and 100°-40°-40° involved.

P.S.(2) Note to Gianlino: Thanks for those kind words, problems, involving angles, multiple of π/9, are thrilling indeed. My congratulations for Your solution too!

BC/sin(40 = AC/sin(100) -->

AC = BC*[sin(40)/sin(100)] = BC*1.53208889.

After extending the line, we can also see that

BC/sin BDC = CD/sin(100) --> sin BDC = (BC/CD)*sin(100)

The Law of Cosines tells us that

CD^2 = BC^2 + BD^2 - 2(BC)(BD)*cos(100) -->

CD^2 = BC^2 + AC^2 - 2(BC)(AC)*cos(100) -->

CD^2 = BC^2 + (BC*1.53208889)^2 - 2(BC)(BC*1.53208889)*cos(100) -->

CD^2 = BC^2 + 2.347296355BC^2 + 0.53208889*BC^2 -->

CD^2 = BC^2 * 3.879385241 -->

CD/BC = 1.969615506

sin BDC = (BC/CD)*sin(100) = (1/1.969615506)*sin(100) -->

sin BDC = 1/2 (check it with a calculator) -->

BDC = 30 degrees.

Now...I did all that calculation and then saw that I was looking at the *wrong* angle. I solved BDC but you wanted ACD. No problem.

BAC = 40 degrees and we extended BA out to D, so CAD = 180 - 40 = 140 degrees.

--> ACD = 180 - 140 - 30 -->

ACD = 10 degrees.

edit: ah so. hrm...I will see what I can come up with.

its about 18.677 degrees

other angles

its about 18.677 degrees

other angles

CAD = 140

ADC = 21.32

ACD = 18.677 (Required)

Source(s):

http://en.wikipedia.org/wiki/Triangle#Further_form...

(for formulas of triangle)

that was a lengthy one,,,

its about 18.677 degrees

other angles

CAD = 140

ADC = 21.32

ACD = 18.677 (Required)