Let $C$ is at a circle. $[AB]$ is a diameter this circle. $D$ is a point at $[AB]$. Perpendicular from $C$ to $[AB]$'s foot on the $[AB]$ is $E$, perpendicular from $A$ to $[CD]$'s foot on the $[CD]$ is $F$. Prove that, $$[DC][FC]=[BD][EA]$$
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Tags: geometry
Let $C$ is at a circle. $[AB]$ is a diameter this circle. $D$ is a point at $[AB]$. Perpendicular from $C$ to $[AB]$'s foot on the $[AB]$ is $E$, perpendicular from $A$ to $[CD]$'s foot on the $[CD]$ is $F$. Prove that, $$[DC][FC]=[BD][EA]$$