**Introduction to Constructions**

**Linear Pair axiom**

- If a ray stands on a line then the adjacent angles form a linear pair of angles.
- If two angles form a linear pair, then uncommon arms of both the angles form a straight line.

**Angle Bisector**

Construction of an Angle bisector

Suppose we want to draw the angle bisector of âˆ ABC we will do it as follows:

- Taking B as centre and any radius, draw an arc to intersect AB and BC to intersect at D and E respectively.
- Taking D and E as centres and with radius more than DE/2, draw arcs to intersect each other at a point F.
- Draw the ray BF. This ray BF is the required bisector of the âˆ ABC.

**Perpendicular Bisector**

Construction of a perpendicular bisector

Steps of construction of a perpendicular bisector on the line segment AB:

- Take A and B as centres and radius more than AB/2 draw arcs on both sides of the line.
- Arcs intersect at the points C and D. Join CD.
- CD intersects AB at M. CMD is the required perpendicular bisector of the line segment AB.

Proof of validity of construction of a perpendicular bisector

Proof of the validity of construction of the perpendicular bisector:

Î”DAC and Î”DBC are congruent by SSS congruency. (âˆµ AC = BC, AD = BD and CD = CD)

âˆ ACM and âˆ BCMare equal (cpct)

Î”AMC and Î”BMC are congruent by SAS congruency. (âˆµ AC = BC, âˆ ACM = âˆ BCM and CM = CM)

AM = BM and âˆ AMC = âˆ BMC (CPCT)

âˆ AMC + âˆ BMC = 180^{0} (Linear Pair Axiom)

âˆ´ âˆ AMC = âˆ BMC = 90^{âˆ˜}

Therefore, CMD is the perpendicular bisector.

**Constructing Angles**

Construction of an Angle of 60 degrees

Steps of construction of an angle of 60 degrees:

- Draw a ray QR.
- Take Q as the centre and some radius draw an arc of a circle, which intersects QR at a point Y.
- Take Y as the centre with the same radius draw an arc intersecting the previously drawn arc at point X.
- Draw a ray QP passing through X
- âˆ PQR=60
^{âˆ˜}

Proof for the validity of construction of an Angle of 60 degrees

Proof for the validity of construction of the 60^{0} angle:

Join XY

XY = XQ = YQ (By construction)

âˆ´â–³XQY is an equilateral triangle.

Therefore, âˆ XQY = âˆ PQR = 60^{0}

**Triangle Constructions**

Construction of triangles

At least three parts of a triangle have to be given for constructing it but not all combinations of three parts are sufficient for the purpose.

Therefore a unique triangle can be constructed if the following parts of a triangle are given:

- two sides and the included angle is given.
- three sides are given.
- two angles and the included side is given.
- In a right triangle, hypotenuse and one side are given.
- If two sides and an angle (not the included angle) are given, then it is not always possible to construct such a triangle uniquely.

Given base, base angle and sum of other two sides

**Steps for construction of a triangle given base, base angle, and the sum of other two sides:**

- Draw the base BC and at point B make an angle say XBC equal to the given angle.
- Cut the line segment BD equal to AB + AC from ray BX.
- Join DC and make an angle DCY equal to âˆ BDC.
- Let CY intersect BX at A.
- ABC is the required triangle.

Given base(BC), base angle(ABC) and AB-AC

**Steps of construction of a triangle given base(BC), base angle(****âˆ ****ABC) and difference of the other two sides (AB-AC):**

- Draw base BC and with point B as the vertex make an angle XBC equal to the given angle.
- Cut the line segment BD equal to AB â€“ AC(AB > AC) on the ray BX.
- Join DC and draw the perpendicular bisector PQ of DC.
- Let it intersect BX at a point A. Join AC.
- Then â–³ABC is the required triangle.

Proof for validation for Construction of a triangle with given base, base angle and difference between two sides

Validation of the steps of construction of a triangle with given base, base angle and difference between two sides

- Base BC and âˆ B are drawn as given.
- Point A lies on the perpendicular bisector of DC. So, AD = AC.
- BD = AB â€“ AD = AB â€“ AC (âˆµ AD = AC).
- Therefore ABC is the required triangle.

Given base (BC), base angle (ABC) and AC-AB

**Steps of construction of a triangle given base (BC), base angle (****âˆ ****ABC) and difference of the other two sides (AC-AB):**

- Draw the base BC and at point B make an angle XBC equal to the given angle.
- Cut the line segment BD equal to AC â€“ AB from the line BX extended on the opposite side of line segment BC.
- Join DC and draw the perpendicular bisector, say PQ of DC.
- Let PQ intersect BX at A. Join AC.
- â–³ABC is the required triangle.

Given perimeter and two base angles

**Steps of construction of a triangle with given perimeter and two base angles.**

- Draw a line segment, say GH equal to BC + CA + AB.
- Make angles XGH equal to âˆ B and YHG equal to âˆ C, where angle B and C are the given base angles.
- Draw the angle bisector ofâˆ XGH and âˆ YHG. Let these bisectors intersect at a point A.
- Draw perpendicular bisectors PQ of AG and RS of AH.
- Let PQ intersect GH at B and RS intersect GH at C. Join AB and AC
- â–³ABC is the required triangle.

Proof for validation for Construction of a triangle with given perimeter and two base angles

Validating the steps of construction of a triangle with given perimeter and two base angles:

- B lies on the perpendicular bisector PQ of AG and C lies on the perpendicular bisector RS of AH. So, GB = AB and CH = AC.
- BC + CA + AB = BC + GB + CH = GH (âˆµ GB = AB and CH = AC)
- âˆ BAG = âˆ AGB (âˆµÎ”AGB, AB = GB)
- âˆ ABC = âˆ BAG + âˆ AGB = 2âˆ AGB = âˆ XGH
- Similarly, âˆ ACB =âˆ YHG