Solution a) Prove: SB ⊥ BC Recall that ABCD is a rhombus by assumption then uur uuur uur uuur SD + AB − SA2 − BD SB.BC = SB AD = 2.SB AD We get Since (1) SA =SD ( by assumption) ( 2) AB = BD ABD is an equiangular traingle ( ) It follows from (1) and (2) that Thus SB ⊥ uur uuur uur uuur r SB AD = SB.BC = BC b) tan=? On the plane (ABCD), let O=AC ∩ BD