Question

$O$ is the center of circle.$AB$ and $AC$ are tangent drawn from $A$ and $BA$ perpendicular to $AC$. Prove that $BACO$is a square.

Answer 1

Given: $O$ is the centre. Now,
$\angle BAC={90}^{\circ }$
$AB=AC$(Tangents drawn from a point onto the circle will be equal in length)
$OB=OC$ (radius)
$\angle ABO=\angle ACO={90}^{\circ }$ (radius is perpendicular to the tangent at the point of contact)
$OB\parallel AC$ (Two lines perpendicular to the same line are parallel to each other).
$AB=OC$ (The distance between two parallel lines in a direction perpendicular to both the lines is constant)
$\Rightarrow AB=AC=OC=OB$
All sides are equal in length
All angles measure ${90}^{\circ }$
Hence it is a square.