**Construction 1**: **How to construct the bisector of an angle?**

If we have to bisect the BOA, then we need to follow these steps to construct the bisector of the angle.

**Step 1:** Take O as the centre and draw an arc by any radius intersecting the rays OA and OB at X and Y respectively.

**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 XY.

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

In the above example, BOA = 80° and OC bisects it in BOC and 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 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, (SSS rule)

So, (CPCT)

Now in triangles CMA and CMB,

AC = BC (As before)

CM = CM (Common)

(Proved above)

Therefore, (SAS rule)

So, AM = BM and CMA = CMB (CPCT)

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

Now we get

CMA = 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 RPQ is the required angle of 60°.

Join AB.

Then, AP = AB = PB (By construction)

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

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