The solution (1)∫ quadrilateral ABCD is rhombic diagonal BD, with side length 13 cm and length 10 cm AC⊥BD.

Let the intersection of BD and AC be e, then BE=ED AE=EC.

∴ EC 2 = BC 2-be 2 BC = 13 BE=5 ∴AC= 12.

(2)S△BCD = BD×EC/2 = 10×5/2 = 25

SABCD=25×2=50

2. Prove that the connecting line EF passes through the point H H∵△ABO?△ADO.

And ∵E and f are AB, respectively, and the midpoint of AD ∴ eh = HF and EF⊥AC.

∴△EHO≌△FHO ∴OE=OF