Step 2: Now take X and Y as the centre and Draw arcs which intersects each other at a point C with radius more than $\left(\frac{1}{2}\right)$ XY.

Step 3: Join OC to draw a ray which is the required bisector of the $\angle$ BOA.

In the above example, $\angle$ BOA = 80° and OC bisects it in $\angle$ BOC and $\angle$ COA which is 40° each.

Construction 2: How to construct the perpendicular bisector of a given line segment?

We have to bisect the given line segment AB.

Step 1: Take A and B as the centres and radius more than $\frac{1}{2}$ of Ab and draw the arcs on both sides.

Step 2: These arcs should intersect each other at C and D. And join CD.

Step 3: Here CD intersects AB at point M so that M is the midpoint of AB and CMD is the required perpendicular bisector of AB.

Join A and B to both C and D to form AC, AD, BC and BD.

In triangles CAD and CBD,

AC = BC (Arcs of equal radii)

AD = BD (Arcs of equal radii)

CD = CD (Common)

Therefore, $\mathrm{\triangle CAD\cong\triangle CBD}$ (SSS rule)

So, $\mathrm{\angle ACM=\angle BCM}$ (CPCT)

Now in triangles CMA and CMB,

AC = BC (As before)

CM = CM (Common)

$\mathrm{\angle ACM=\angle BCM}$ (Proved above)

Therefore, $\mathrm{\triangle CMA\cong\triangle CMB}$ (SAS rule)

So, AM = BM and $\angle$ CMA = $\angle$ CMB (CPCT)

As $\angle$ CMA + $\angle$ CMB = 180° (Linear pair axiom),

Now we get

$\angle$ CMA = $\angle$ CMB = 90°.

Therefore, CMD is the perpendicular bisector of AB.

Construction 3: How to construct an angle of 60° at the initial point of a given ray?

We have to draw an angle of 60° at the given point P.

Step 1: Take P as the centre and draw an arc of any radius which intersects PQ at point B.

Step 2: Now Take B as a centre and draw an arc with the same radius as before which intersects the previous arc at point A.

Step 3: Now draw a ray PR which passé through Point A and the $\angle$ RPQ is the required angle of 60°.

Join AB.

Then, AP = AB = PB (By construction)

Therefore, ∆ ABP is an equilateral triangle and the $\angle$ APB, which is the same as $\angle$ RPQ is equal to 60°.

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