If the diagonals of a quadrilateral divide each other proportionally then it is a

In the given figure, $ABCD$ is a trapezium whose diagonals $AC$ and $BD$ intersect at $O$ such that $OA=\left(3x-1\right) \mathrm{cm}, OB=\left(2x+1\right) \mathrm{cm}, OC=\left(5x-3\right) \mathrm{cm}$ and $OD=\left(6x-5\right) \mathrm{cm}$. Then, $x=?$

The line segments joining the midpoints of the adjacent sides of a quadrilateral forms

If the bisector of an angle bisects the opposite side, then the triangle is 

In $∆ABC$, it is given that $\frac{AB}{AC}=\frac{BD}{DC}$. If $\angle B=70°$ and $\angle C=50°$, then $\angle BAD=?$

In $∆ABC,DE\parallel BC$, so that $AD=2.4 \mathrm{cm}, AE=3.2 \mathrm{cm}$ and $EC=4.8 \mathrm{cm}$. Then, $AB=?$

In a $∆ABC$, if $DE$ is drawn parallel to $BC$, cutting $AB$ and $AC$ at $D$ and $E$ respectively such that $AB=7.2 \mathrm{cm}, AC=6.4 \mathrm{cm}$ and $AD=4.5 \mathrm{cm}$. Then, $AE=?$

In $∆ABC, DE\parallel BC$ so that $AD=\left(7x-4\right) \mathrm{cm}, AE=\left(5x-2\right) \mathrm{cm}, DB=\left(3x+4\right) \mathrm{cm}$and $EC=3x \mathrm{cm}$. Then, we have

In $△\mathrm{ABC}, \mathrm{DE}\left|\right|\mathrm{BC}$ such that $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{5}$. If $\mathrm{AC}=5.6 \mathrm{cm}$, then $\mathrm{AE}=$