$ABCD$ is a square with $AB=1$. $F$ lies on segment $CD$ and segment $BF$ intersects with segment $AC$ at point $E$ such that $EF=DF$. What is the length of $BE$?
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Solution:
Let $G$ be a point on segment $CD$ such that $EG$ is orthogonal to $CD$. Let $H$ be a point on segment $AD$ such that $EH$ is orthogonal to $EG$. We see that $\angle HED=\angle DEF=\angle FEG$, so they are all $30^\circ$.
Extend $D$ to $I$ away from $C$ on line $CD$ such that $ID=BE$. We have $\angle BIC=30^\circ$, so $BE=ID=IC-DC=\sqrt{3}-1$.
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