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Point $E$ lies on the extended side $AD$ of parallelogram $ABCD$. $BE$ intersects $CD$ at $F$. Show that (i) $\Delta DFE \sim \Delta CFB$ (ii) $\Delta AEB \sim \Delta CBF$.
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Solution:
(i) In $\Delta DFE$ and $\Delta CFB$
$\angle 5 = \angle 3$ (Alternate Interior Angle)
$\angle 1 = \angle 2$ (Alternate Interior Angle)
$\therefore$ By AA similarity criterion, $\Delta DFE \sim \Delta CFB$
(ii) In $\Delta AEB$ and $\Delta CBF$
$\angle 1 = \angle 2$ (Alternate Interior Angle)
$\angle 4 = \angle 3$ (Opposite angles of a parallelogram)
$\therefore$ By AA similarity criterion, $\Delta AEB \sim \Delta CBF$
(i) In $\Delta DFE$ and $\Delta CFB$
$\angle 5 = \angle 3$ (Alternate Interior Angle)
$\angle 1 = \angle 2$ (Alternate Interior Angle)
$\therefore$ By AA similarity criterion, $\Delta DFE \sim \Delta CFB$
(ii) In $\Delta AEB$ and $\Delta CBF$
$\angle 1 = \angle 2$ (Alternate Interior Angle)
$\angle 4 = \angle 3$ (Opposite angles of a parallelogram)
$\therefore$ By AA similarity criterion, $\Delta AEB \sim \Delta CBF$