Arafat wrote:I could solve the 1st part only....can anyone explain the another one.....

Here is my ans ( I'm not sure wheter i'm correct )

**2nd Part **
$\angle ACB = 30^{\circ}$

$\angle ABC = 90^{\circ}$ ( Since it is a right angled triangle )

$\angle BAC + 30^{\circ} + 90^{\circ} = 180^{\circ}$

$\angle BAC= 60^{\circ}$

$\angle ABD= \angle ABC - \angle CBD\; \; (\angle CBD = \angle BCD=30^{\circ}) =90^{\circ}-30^{\circ}=60^{\circ}$

$\angle BDA+\angle DAB+\angle ABD= 180^{\circ}$

$\angle BDA = 60^{\circ}$

$\angle BDA =\angle EDF= 60^{\circ}$

$\angle FGC + \angle FCG + \angle GFC = 180^{\circ}$

$90^{\circ} + 30^{\circ} + \angle GFC = 180^{\circ}$

$\angle GFC = 60^{\circ}$

$\angle GFC = \angle DFE=60^{\circ}$

$\angle EDF +\angle DFE + \angle DEF = 180^{\circ}$

$60^{\circ} + 60^{\circ} + \angle DEF = 180^{\circ}$

$\angle DEF= 60^{\circ}$

All three angles are equal so $\triangle DEF$ is isosceles { Solved }

I hope I'm right.

Anyone plz check my ans