MathDB

2004.11.4

Part of Kyiv City MO Seniors 2003+ geometry

Problems(1)

<EFA=< PQB_1 in rect. parallelepiped (Kyiv City Olympiad 2004 11.4)

Source:

6/28/2021
Given a rectangular parallelepiped ABCDA1B1C1D1ABCDA_1B_1C_1D_1. Let the points EE and FF be the feet of the perpendiculars drawn from point AA on the lines A1DA_1D and A1CA_1C, respectively, and the points PP and QQ be the feet of the perpendiculars drawn from point B1B_1 on the lines A1C1A_1C_1 and A1CA_1C, respectively. Prove that ∠EFA=∠PQB1\angle EFA = \angle PQB_1
geometryequal angles3D geometryparallelepiped