80-80-20 Triangle–Case I: 50-60
Michael N. Fried
December 6, 2016
This is a well-known problem with many solution and several variations.
Let ABC be an isosceles triangle whose base angles are both 80◦ . Let CD be
drawn at a 60◦ angle to BC and BE at a 50◦ angle. Join DE what is the angle
D2 ?
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Draw BH at a 60◦ angle to BC. Let BH intersect CD at F : join DH and F E.
First, note that angle E4 is 50◦ , so that EC = BC = CF . Therefore, B, F, E
lie on a circle with center C. Hence, since 6 F CB = 60◦ , it follows that
6 F EB = 1 60◦ = 30◦ . Angle G1 = F2 = 70◦ , so F3 = 80◦ . Angle F1 = 60◦ , so
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F2 = 40◦ . But also H2 = 40◦ , so HE = F E, and, since F D = DH (DHF is an
equilateral triangle), F EHD is a kite, so that DE and F H are perpendicular,
and DE bisects the angles at D and E. Therefore, E2 = E1 = 21 100◦ = 50◦ and
D2 = D1 = 12 60◦ = 30◦ .
From here, the first variation, namely, where CD is drawn 60◦ to BC, BJ
at 70◦ , and J1 must be found, is fairly easy.
Draw BE again at 50◦ ro BC and join F E. From before, we know that
D1 = 30◦ . But, also J2 = 30◦ , so D, J, E, F are concyclic. Hence, J1 = E2 .
Again, observe that E4 = F1 = 50◦ , so that F, E, C, B are concylic. Hence
F2 = 50◦ . But E2 = F2 − D1 = 50◦ − 30◦ = 20◦ , so that also J1 = 20◦ .
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