Let $ABCD$ be a trapezium with $AD$ parallel to $BC$. Assume there is a point $M$ in the interior of the segment $BC$ such that $AB=AM$ and $DC=DM$. Then,the ratio of the area of the trapezium to the area of $\triangle AMD$ is

Suppose $ABOC$ is a rhombus in the first quadrant with $O$ being the origin. If the vertices $B$ and $C$ of $\triangle ABC$ lie respectively on $y=\frac{4}{3}x$ and $y=0$,and the side $BC$ passes through $\left(\frac{2}{3}, \frac{2}{3}\right)$,then the mid-point of $BC$ is

Without using the distance formula,show that the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ are the vertices of a parallelogram.

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to:

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

Let the circumcentre of a triangle with vertices $A(a, 3)$,$B(b, 5)$,and $C(a, b)$,where $ab > 0$,be $P(1, 1)$. If the line $AP$ intersects the line $BC$ at the point $Q(k_{1}, k_{2})$,then $k_{1} + k_{2}$ is equal to.