Question
In Fig., $\text{DEFG}$ is a square in a triangle $\text{ABC}$ right angled at $A$ . Prove that
$i. \triangle AGF \sim \triangle DBG$
$ii. \triangle AGF \sim \triangle EFC$
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Answer

$GF \| DE\ (\text{DEFG}$ is square$)$
$\therefore \angle AGF=\angle ABC \ ($Corresponding angles$)$
$\therefore \angle A=\angle GDB=90^{\circ}$
$\therefore \angle AGF \sim \angle DBG \ ($By $AA$ similarity$)$
Again $\text{DEFG}$ being a square $\angle AFG =\angle ACB\  ($corresponding angles$)$
$\therefore \angle A=\angle CEF\ ($each $ 90^{\circ})$
$\angle AGF \sim \angle EFC\ ($By $AA$ similarity$)$
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