Question
Construct with ruler and compass, angle of measure $90^\circ .$
✓
Answer
$i.\ $Draw any line $PQ$ and take a point $O$ on it.
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with A as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with B as centre, draw an arc which cuts the first arc at $C.$
$v.\ $Join $OB$ and $OC.$
$vi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle COB.$ Label the points of intersection as $D$ and $E.$
$vii.\ $With $E$ as centre, draw $($in the interior of $\angle COB)$ an arc where radius is more than half the length $ED.$
$viii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle COB.$ Let the two arcs intersect at $F.$ Join $\overline{OF}$. Then, $\overline{OF}$ is the bisector of $\angle COB,$ i.e., $\angle COF = \angle FOB.$ Now, $\angle FOQ = 90^\circ .$
$ii.\ $Place the pointer of the compasses at $O$ and draw an arc of convenient radius which cuts the line at $A.$
$iii.\ $Without disturbing the radius on the compasses, draw an arc with A as centre which cuts the first arc at $B.$
$iv.\ $Again without disturbing the radius on the compasses and with B as centre, draw an arc which cuts the first arc at $C.$
$v.\ $Join $OB$ and $OC.$
$vi.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle COB.$ Label the points of intersection as $D$ and $E.$
$vii.\ $With $E$ as centre, draw $($in the interior of $\angle COB)$ an arc where radius is more than half the length $ED.$
$viii.\ $With the same radius and with $D$ as centre, draw another arc in the interior of $\angle COB.$ Let the two arcs intersect at $F.$ Join $\overline{OF}$. Then, $\overline{OF}$ is the bisector of $\angle COB,$ i.e., $\angle COF = \angle FOB.$ Now, $\angle FOQ = 90^\circ .$
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