Two similar triangles ABC and EBD. Two line segments meet at point F. Prove that ABFC and EBDF are cyclic quadrilaterals.
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Solution
ΔABC∽ΔEBD
AB:BE=BC:BD
ΔABE∽ΔCBD
∠BAE=∠BCD, ABFC cyclic
∠AEB=∠CDB, EBDF cyclic