MathDB
AD + DX - (BC + CX) = 8 , trapezoid ABCD

Source: Mathcenter Contest / Oly - Thai Forum 2008 R2 p4 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 11, 2022
geometrytrapezoid

Problem Statement

The trapezoid ABCDABCD has sides ABAB and CDCD that are parallel DAB^=6∘\hat{DAB} = 6^{\circ} and ABC^=42∘\hat{ABC} = 42^{\circ}. Point XX lies on the side ABAB , such that AXD^=78∘\hat{AXD} = 78^{\circ} and CXB^=66∘\hat{CXB} = 66^{\circ}. The distance between ABAB and CDCD is 11 unit . Prove that AD+DX−(BC+CX)=8AD + DX - (BC + CX) = 8 units.
(Heir of Ramanujan)
Back to ProblemsView on AoPS