<pre>
 1. Let K is a point on side AB (different from B) such that CK=CB.
 2. Let L is a point on AC (different from C) such that KL=KC.
 3. Let M is a point on AB (different from K) such that LM=LK.

Now I prove that M=D.

Triangle KCB is isosceles, angle CBK<code>80  </code>> CKB<code>80 and BCK</code>20.

Triangle LKC is isosceles, angle LCK<code>80-20</code>60 => this triangle is
equilateral, LKC<code>CLK</code>60, LC<code>KL</code>KC.

Triangle KLM is isosceles, angle LKM <code> 180-80-60 </code> 40  <code>> LMK</code>40 and MLK=100.

In triangle MLC angle L <code> 100+60</code>160 and LC<code>LM   </code>> LCM<code>LMC</code>10.

So, LCM<code>LCD (</code>80-70) and M in AB <code>> M</code>D ====
====

Consider LEDB. Angle ELD is equal to 180-160<code>20 degrees, i.e. </code> angle EBD.
So points L, B, D and E lies on one circumference ====
====
(From DLE<code>20 and DAE</code>20 we have another proof that AD<code>LD </code> ... = BC ====)
====

We get that angle EDL<code>EBL, DEB</code>DLB and so on.

Now find angle EDC:

EDC<code>LDC+LDE</code>10+LDE<code>10+LBE</code>10+(CBE-CBL)=70-CBL.

But triangle BCL is isosceles <code>> CBL</code>(180-80)/2 = 50;
EDC<code>70-50</code>20.

      Konstantin Knop,
         St.Petersburg, Russia.
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