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=80 => CKB=80 and BCK=20.

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

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

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

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

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

We get that angle EDL=EBL, DEB=DLB and so on.

Now find angle EDC:

EDC=LDC+LDE=10+LDE=10+LBE=10+(CBE-CBL)=70-CBL.

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

Konstantin Knop,

St.Petersburg, Russia.

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