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)
Now in triangles CMA and CMB,
AC = BC (As before)
CM = CM (Common)
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°.
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°.