In trapezium $ABCD$,$AB \parallel CD$. If $\angle A = 70^{\circ}$ and $\angle B = 85^{\circ}$,then $\angle D = \ldots$ (in $^{\circ}$)

Three angles of a quadrilateral are $75^{\circ}, 90^{\circ}$ and $75^{\circ}$. The fourth angle is (in $^{\circ}$)

$A$ square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.

Prove that the diagonals of a square are equal and bisect each other at right angles.