Geometry puzzle (22) Asked: find angle x in the figure below. This is not simple, so first we calculate some angles: We notice two isosceles triangles: ΔABD and ΔBCS. Now we apply some intuition. In the case of isosceles triangles, the top angle is on the perpendicular bisector of the base. The intersection of two perpendicular bisectors is the center of the circumscribed circle of the triangle. We construct the circumscribed circle of ΔBCD SM is perpendicular bisector of BC. TM is perpendicular bisector of BD. DP is perpendicular bisector of AB. Quadrilateral TBMD is a rhombus because MB = MD BT = DT diagonals BD and MT are perpendicular. This implies that DT and MB are parallel so LABM = 90. AB is tangent of the circle. LABD = LBCD because they span the same circle arc. x = LACD = 70 - 30 = 40.