As you have learnt how to construct a circle, the perpendicular bisector of a line segment, angles of `30°, 45°, 60°,` `90°` and `120°`, and the bisector of a given angle, without giving any justification for these constructions.
Here, you will construct some of these, with reasoning behind, why these constructions are validtions
`color{blue} text(Construction 11.1 :) "To construct the bisector of a given angle."`
Given an angle ABC, we want to construct its bisector.
`bb text(Steps of Construction :)`
1. Taking B as centre and any radius, draw an arc to intersect the rays `BA` and `BC`, say at `E` and `D` respectively [see Fig.11.1(i)].
2. Next, taking `D` and `E` as centres and with the radius more than `1/2` `DE`, draw arcs to intersect each other, say at `F`.
3. Draw the ray `BF` [see Fig.11.1(ii)]. This ray `BF` is the required bisector of the angle `ABC`.
Let us see how this method gives us the required angle bisector.
`DF` and `EF`.
In triangles `BEF` and `BDF`,
`BE = BD` (Radii of the same arc)
`EF = DF` (Arcs of equal radii)
`BF = BF` (Common)
Therefore, `Delta BEF ≅ Delta BDF` (SSS rule)
This gives `∠EBF = ∠ DBF (CPCT)`
`color {blue} text(Construction) 11.2 : "To construct the perpendicular bisector of a given line segment."`
Given a line segment AB, we want to construct its perpendicular bisector.
`color {blue} text(Steps of Construction) :`
1. Taking A and B as centres and radius more than `1/2 AB`, draw arcs on both sides of the line segment `AB` (to intersect each other).
2. Let these arcs intersect each other at `P` and `Q`. Join `PQ` (see Fig.11.2).
3. Let `PQ` intersect `AB` at the point `M`. Then line `PMQ` is the required perpendicular bisector of `AB`.
Let us see how this method gives us the perpendicular bisector of `AB`.
Join `A` and `B` to both `P` and `Q` to form `AP, AQ, BP` and `BQ`.
In triangles `PAQ` and `PBQ`,
`AP = BP` (Arcs of equal radii)
`AQ = BQ` (Arcs of equal radii)
`PQ = PQ` (Common)
Therefore, `Delta PAQ ≅ Delta PBQ` (SSS rule)
So, `∠ APM = ∠ BPM (CPCT)`
Now in triangles `PMA` and `PMB`,
`AP = BP` (As before)
`PM = PM` (Common)
`∠ APM = ∠ BPM` (Proved above)
Therefore, `Delta PMA ≅ Delta PMB` (SAS rule)
So, `AM = BM` and `∠ PMA = ∠ PMB (CPCT)`
As `∠ PMA + ∠ PMB = 180°` (Linear pair axiom),
we get
`∠ PMA = ∠ PMB = 90°.`
Therefore, `PM`, that is, `PMQ` is the perpendicular bisector of `AB.`
`color{blue} text(Construction 11.3 :)"To construct an angle of 600 at the initial point of a given ray."`
Let us take a ray `AB` with initial point A [see Fig. 11.3(i)]. We want to construct a ray `AC` such that `∠ CAB = 60°`. One way of doing so is given below.
`color{blue}bb text( Steps of Construction :)`
1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.
2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.
3. Draw the ray AC passing through E [see Fig 11.3 (ii)]. Then `∠ CAB` is the required angle of `60°.` Now, let us see how this method gives us the required angle of `60°.`
Join `DE`.
Then, `AE = AD = DE` (By construction)
Therefore, `Delta EAD` is an equilateral triangle and the `∠ EAD`, which is the same as `∠ CAB` is equal to `60°`.
As you have learnt how to construct a circle, the perpendicular bisector of a line segment, angles of `30°, 45°, 60°,` `90°` and `120°`, and the bisector of a given angle, without giving any justification for these constructions.
Here, you will construct some of these, with reasoning behind, why these constructions are validtions
`color{blue} text(Construction 11.1 :) "To construct the bisector of a given angle."`
Given an angle ABC, we want to construct its bisector.
`bb text(Steps of Construction :)`
1. Taking B as centre and any radius, draw an arc to intersect the rays `BA` and `BC`, say at `E` and `D` respectively [see Fig.11.1(i)].
2. Next, taking `D` and `E` as centres and with the radius more than `1/2` `DE`, draw arcs to intersect each other, say at `F`.
3. Draw the ray `BF` [see Fig.11.1(ii)]. This ray `BF` is the required bisector of the angle `ABC`.
Let us see how this method gives us the required angle bisector.
`DF` and `EF`.
In triangles `BEF` and `BDF`,
`BE = BD` (Radii of the same arc)
`EF = DF` (Arcs of equal radii)
`BF = BF` (Common)
Therefore, `Delta BEF ≅ Delta BDF` (SSS rule)
This gives `∠EBF = ∠ DBF (CPCT)`
`color {blue} text(Construction) 11.2 : "To construct the perpendicular bisector of a given line segment."`
Given a line segment AB, we want to construct its perpendicular bisector.
`color {blue} text(Steps of Construction) :`
1. Taking A and B as centres and radius more than `1/2 AB`, draw arcs on both sides of the line segment `AB` (to intersect each other).
2. Let these arcs intersect each other at `P` and `Q`. Join `PQ` (see Fig.11.2).
3. Let `PQ` intersect `AB` at the point `M`. Then line `PMQ` is the required perpendicular bisector of `AB`.
Let us see how this method gives us the perpendicular bisector of `AB`.
Join `A` and `B` to both `P` and `Q` to form `AP, AQ, BP` and `BQ`.
In triangles `PAQ` and `PBQ`,
`AP = BP` (Arcs of equal radii)
`AQ = BQ` (Arcs of equal radii)
`PQ = PQ` (Common)
Therefore, `Delta PAQ ≅ Delta PBQ` (SSS rule)
So, `∠ APM = ∠ BPM (CPCT)`
Now in triangles `PMA` and `PMB`,
`AP = BP` (As before)
`PM = PM` (Common)
`∠ APM = ∠ BPM` (Proved above)
Therefore, `Delta PMA ≅ Delta PMB` (SAS rule)
So, `AM = BM` and `∠ PMA = ∠ PMB (CPCT)`
As `∠ PMA + ∠ PMB = 180°` (Linear pair axiom),
we get
`∠ PMA = ∠ PMB = 90°.`
Therefore, `PM`, that is, `PMQ` is the perpendicular bisector of `AB.`
`color{blue} text(Construction 11.3 :)"To construct an angle of 600 at the initial point of a given ray."`
Let us take a ray `AB` with initial point A [see Fig. 11.3(i)]. We want to construct a ray `AC` such that `∠ CAB = 60°`. One way of doing so is given below.
`color{blue}bb text( Steps of Construction :)`
1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.
2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.
3. Draw the ray AC passing through E [see Fig 11.3 (ii)]. Then `∠ CAB` is the required angle of `60°.` Now, let us see how this method gives us the required angle of `60°.`
Join `DE`.
Then, `AE = AD = DE` (By construction)
Therefore, `Delta EAD` is an equilateral triangle and the `∠ EAD`, which is the same as `∠ CAB` is equal to `60°`.