Basic Constructions

Construction 1How to construct the bisector of an angle?

If we have to bisect the \angle 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  \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 3How 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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