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(A) In $\Delta XYZ$,the ratio of angles is $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$.Let the angles be $2k, 3k, 5k$.Since the sum of angles i

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$\angle D$

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(A) In $\Delta XYZ$,the ratio of angles is $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$.Let the angles be $2k, 3k, 5k$.Since the sum of angles in a triangle is $180^{\circ}$,we have $2k + 3k + 5k = 180^{\circ}$,which gives $10k = 180^{\circ}$,so $k = 18^{\circ}$.Thus,$m \angle X = 2(18^{\circ}) = 36^{\circ}$,$m \angle Y = 3(18^{\circ}) = 54^{\circ}$,and $m \angle Z = 5(18^{\circ}) = 90^{\circ}$.The correspondence $XYZ \leftrightarrow EFD$ is a similarity,which implies $\Delta XYZ \sim \Delta EFD$.Therefore,the corresponding angles are equal: $m \angle X = m \angle E$,$m \angle Y = m \angle F$,and $m \angle Z = m \angle D$.Since $m \angle Z = 90^{\circ}$,it follows that $m \angle D = 90^{\circ}$.Thus,$\angle D$ is a right angle.

If $\Delta XYZ \sim \Delta DEF$ for the correspondence $XYZ \leftrightarrow EFD$.
If $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$,then in $\Delta DEF$,$\ldots \ldots$ is a right angle.

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If $\Delta XYZ \sim \Delta DEF$ for the correspondence $XYZ \leftrightarrow EFD$.If $m \angle X : m \angle Y : m \angle Z = 2 : 3 : 5$,then in $\Delta DEF$,$\ldots \ldots$ is a right angle.