Question
Construct with ruler and compass, angle of measure $135^\circ .$
✓
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 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 whose 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, $\angle FOQ = 90^\circ .$
$ix.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POF.$ Label the points of intersection as $G$ and $H.$
$x.\ $With $H$ as centre draw $($in the interior of $\angle POF)$ an arc whose radius is more than half the length $HG.$
$xi.\ $With the same radius and with $G$ as centre, draw another arc in the interior of $\angle POF.$ Let the two arcs intersect at $I.$ Join $OI.$ Then, $\overline {OI}$ is the bisector of $\angle POF,$ i.e. $\angle GOI = \angle IOF.$ Now, $\angle IOQ = 135^\circ .$
$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 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 whose 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, $\angle FOQ = 90^\circ .$
$ix.\ $With $O$ as centre and using compasses, draw an arc that cuts both rays of $\angle POF.$ Label the points of intersection as $G$ and $H.$
$x.\ $With $H$ as centre draw $($in the interior of $\angle POF)$ an arc whose radius is more than half the length $HG.$
$xi.\ $With the same radius and with $G$ as centre, draw another arc in the interior of $\angle POF.$ Let the two arcs intersect at $I.$ Join $OI.$ Then, $\overline {OI}$ is the bisector of $\angle POF,$ i.e. $\angle GOI = \angle IOF.$ Now, $\angle IOQ = 135^\circ .$
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