Question

ABCD is a square; X is the mid-point of AB and Y the mid-point of BC.Hence, DX is perpendicular to AY.

If the above statement is true then mention answer as 1, else mention 0 if false.

Answer 1

$AB=BC$ (ABCD is a square)
$\therefore \frac{1}{2}AB=\frac{1}{2}BC$
$\therefore AX=BY$ ...(X and Y are mid points of AB and BC respectively)
In $△AXD$ and $△ABY$,
$\angle DAX=\angle ABY$ (Each ${90}^{\circ }$)
$AX=BY$ (Proved above)
$AD=AB$ (ABCD is a square)
Thus, $△AXD\cong △BYA$ ....($SAS$ test)
$\angle AXD=\angle BYA$ (By cpct)

i.e $\angle AXO=\angle OYB$ ...(I)
Now, In quadrilateral XOYB,
Sum of angles = 360
$\angle XOY+\angle OYB+\angle YBX+\angle BXO={360}^o$
$\angle XOY+\angle OYB+90+180-\angle AXO={360}^o$
But $\angle OYB=\angle AXO$ (From I)
hence, $\angle XOY={90}^{\circ }$
$DX\perp AY$