Question

The base angles of some isosceles triangle are 80°. Point D is on segment AC and point E is on AB such that ∠DBC = 60 and ∠ECB 50 . Find the angle EDB.

Answer 1

Mark K on AC such that KBC = 20°. Draw KB and KE. BEC = ECB, and so BEC is isosceles with BE = BC.

BKC = BCK, and so BKC is isosceles with BK = BC. Therefore BE = BK. EBK = 60°, and so EBK is equilateral.

BDK = DBK = 40° and so BDK is isosceles, with KD = KB = KE.

So KDE is isosceles, with EKD = 40°, since EKC = 140°.

Thus, EDK = 70° (triangle BDK is isosceles EKD = 40°, (180 - 40) / 2 = 70°), yielding EDB =EDK-BDK=70°-40°=30°.