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Fix $H,F$
$F'$ and $F'_1$ are reflections of $F$ across $AB , CD$ respectively .
$F'_2$ is reflection of $F'_1$ across $AD$ .
Now $h_{min}=F'F'_2=2\sqrt2$
$F'$ and $F'_1$ are reflections of $F$ across $AB , CD$ respectively .
$F'_2$ is reflection of $F'_1$ across $AD$ .
Now $h_{min}=F'F'_2=2\sqrt2$