The perimeter of rhombus $ABCD$ is $116$. If $AC = 42$, find $BD$.

In $Fig.$,line segment $DF$ intersects the side $AC$ of a triangle $ABC$ at the point $E$ such that $E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$. Prove that
$\frac{BD}{CD} = \frac{BF}{CE}$

In a quadrilateral $ABCD$,$\angle A + \angle D = 90^{\circ}$. Prove that $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.

State whether the following statement is true or false. Justify your answer.
$\triangle ABC$ with vertices $A(-2, 0), B(2, 0)$ and $C(0, 2)$ is similar to $\triangle DEF$ with vertices $D(-4, 0), E(4, 0)$ and $F(0, 4)$.

In $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$. If $PQ = 2 \sqrt{6}$,then $PR = \ldots$

$\square PQRS$ is a rectangle. $PQ + QR = 7$ and the area of $\square PQRS$ is $12$. Then,$PR = \ldots$