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Geometry is a study of shapes. The study involves pictorial views and how to construct the different types of geometric shapes. Also there are some geometrical aids are needed like drawing angle bisectors, perpendicular bisectors etc. Even to draw these, geometrical constructions are used as techniques. Hence, geometric constructions extend to construction of these geometric aids.

To help in the geometric constructions, we certainly be needing the basic geometry tools like scale ruler, protractor, compass, divider etc. It may be noted it is more accurate to use the compass than the scale ruler for for marking a given length.

Let us take a closer look of all the aspects of geometric constructions.

To help in the geometric constructions, we certainly be needing the basic geometry tools like scale ruler, protractor, compass, divider etc. It may be noted it is more accurate to use the compass than the scale ruler for for marking a given length.

Let us take a closer look of all the aspects of geometric constructions.

1) Measures of all the three sides.

2) Measures of any two sides and the 'included' angle.

3) Measure of any one side and measures of any two angles.Let us discuss how a triangle is constructed when the measures of all the three sides are known. Let 'a', 'b' and 'c' are the measures of the three sides. The following diagram shows how the construction is done.

First draw a ray AX which is apparently longer than the measure of 'c'. Now set the compass for a measure of 'c' units. Now without changing the setting of the compass, place the compass needle at A and strike an arc on AX which intersects it at B. Next change the compass setting to 'b' units and draw an arc with center at A. Then change the setting again to 'a' units and draw another arc with center at B. Let these two arcs intersect at point C. Join A to C and B to C. Triangle ABC is the required triangle.

An interesting point may be noted here. Suppose the the sum of 'b' and 'c' is less than or equal to 'a', what happens? In such a case, the arc drawn from A and the arc drawn from B will not intersect at all and they will only be intersecting the line segment AB separately with a gap between the arcs if b + c < a or will only be intersecting AB at the same point if b + c = a. . In other words, a triangle is not possible to construct or does not exist under such condition. Same will be the case if, 'c + a $\leq$ b' or 'a + b $\leq$ c'.

Thus it gives rise to a very important concept that a triangle does not exist if the sum of the measures of any two sides is less than or equal to the measure of the remaining side.

We will next study how to construct a triangle, given the measures of any two sides and the measure of 'included angle'. Let the measure of angle A and the measures 'c' and 'b' are given. Refer to the following diagram.

Let the measure of the angle at vertex A be θ. Draw a ray AX apparently greater than the measure 'c'. With A as center draw an arc for 'c' units to intersect AX at B. Now place the center of the protractor at A and ensure that its 0-line coincides with the line segment AB. Read the angle θ on the protractor and mark that point as P. Remove the protractor, join AP and produce it to Y, such that the line AY is obviously greater than the measure of 'b'. Next place the compass needle at A and draw an arc for 'b' units which intersects AY at C. Join C and B. ABC is the required triangle.

It may be noticed that we have insisted that the given angle must be the 'included angle'. That is, it is angle between the straight lines the measures of which are known. Else, for the same distance of 'b', C can lie anywhere on a circle with center at A and radius of 'b'.

Let us now see how to construct a triangle with the measure of any side and measures of any two angles are known. Unlike the previous case, there are no restrictions on which two angles are given. It is because, once we know any two angles in a triangle, the remaining angle can also be figured out. Let the measure of the side 'c' be known and the angles A (= θ) and B (= α) are known. If A and C or if B and C are known we can always figure out the angle B or angle A. Hence, let us continue with the assumption that measures of angles A and B are known. The construction is explained with the help of the following diagram.

As before, draw a ray AX apparently greater than the measure 'c'. With A as center draw an arc for 'c' units to intersect AX at B. Now place the center of the protractor at A and ensure that its 0-line coincides with the line segment AB. Read the angle θ on the protractor and mark that point as P. Remove the protractor, join AP and produce it to Y, such that the line AY is considerably long. Next place the center of the protractor at B and ensure that its 0-line coincides with the line segment BA. Read the angle α on the protractor and mark that point as Q. Remove the protractor, join BQ and produce it to Z, such that the line BZ is considerably long. Let C be the point of intersection of lines AY and BZ. ABC is the required triangle.

Let AB be the given line segment. With center at A and with a radius which appears to be greater than half of the measure of AB, strike arcs on both sides of the line segment. Now without changing the compass setting, strike two arcs with center at B and again on both sides of the line segments. Let the arcs intersect at point C on one side of the line segment and intersect at D on the other side. Draw a line passing through CD and it is the perpendicular bisector of the line segment AB.

We can furnish the proof as below.

The triangles ACB and ADB are congruent. (The line segments AC, BC, AD, and BD are all congruent by construction and AB is a common side)

Therefore, as per CPCT, angles CAO, CBO, DAO and DBO are all congruent and equal to 'x' (say).

Similarly, the triangles DAC and DBC are congruent. (The line segments AC, BC, AD, and BD are all congruent by construction and DC is a common side)

Therefore, as per CPCT, angles ACO, ADO, BCO and BDO are all congruent and equal to 'y' (say).

Now it is very clear that the triangles ACO and BCO are congruent as per ASA postulate. Hence AO is congruent to BO (CPCT)

Also the angles COA and COB are congruent (CPCT).

But AOB is a straight line as per given condition.Therefore, the angles COA and COB are right angles.

Thus, CO (and hence line CD) is the perpendicular bisector to line segment to AB.

Let us now look into the construction of the internal angle bisector for an angle that is given.

Let AOB is the given angle. From the vertex O, draw two equal arcs intersecting the ray OA at P and intersecting the ray OB at Q. Now with P as center, with a sufficient radius (this radius need not be the same as the radius of the previous arc), draw another arc. Now, with the same compass setting and with Q as center, strike an arc to intersect the arc drawn from P at point C. Draw a ray passing through O and C. The ray OC is the internal angle bisector of the angle AOB.

The proof of this is more simple than the earlier case.

Join PC and join QC.

As per sss postulate, the triangles OPC and OQC are congruent.

Now as CPCT, angle AOC = angle BOC.

Hence the ray OC bisects the angle AOB.

We will learn an important concept on internal angle bisectors. Let CR and CS be the perpendiculars drawn from C on OA and OB. Then, the triangles OCR and OCS are found to be congruent as per AAS postulate, as OC is common and angles ROC and CRO are congruent to angles SOC and CSO respectively. Therefore, CR is congruent to CS. Thus the following conclusion can be drawn in general.

The perpendicular distances on the arms of an angle from any point on the internal angle bisector are congruent.

In figure (i), let A and B are any two points. The line 'l' is the perpendicular bisector of the line segment joining A and B and as per the property of perpendicular bisector , all the points on the line 'l' are equidistant from the points A and B. In other words, infinite number of circles can be drawn through the points A and B. As an example, three circles are shown.

In figure, there are three points A, B and C. Join AB and BC. Draw the perpendicular bisectors 'l' and 'm' of both the line segments which intersect at point O. Since O lies on the perpendicular bisector 'l', points A and B are equidistant from O. Since O lies also on the perpendicular bisector 'm', points B and C are also equidistant from O. In other words O is equidistant from all the three points A, B and C and it is possible to draw a circle through A , B and C with O as center. In fact this is the method of construction of a circle with the given three points. Because two lines can intersect only at one point, it is possible to construct only one circle passing through three given points.

Now, what happens if three points are collinear? In such a case, A, B and C will be in a straight line and the perpendicular bisectors of AB and BC will be parallel which cannot intersect. Therefore, it is not possible to construct a circle passing through three points if they are collinear.

The figure (iii) gives us an important concept. There are four points A, B, C and D. Let the perpendicular bisectors 'l' and 'm' of AB and BC intersect at O. Now it is not probable that the perpendicular bisector 'n' will pass through O. Therefore, for a given set of four points it may not possible to construct a circle passing through all the four points.

The circle that passes through all the three vertices of any triangle is called the circumcircle of the triangle. It is constructed as explained below.

In the above diagram PQR is a scalene triangle. The same concept of constructing a circle passing through three given points is used to construct the circumcircle of a triangle. The vertices are the three given points in this case. That is, the center of the circumcircle C is found by the intersection of the perpendicular bisectors of the sides of the triangle and the circle is constructed by drawing the curve with C as center and with CP (= CQ = CR = R) as radius.

The circle that is constructed inside a triangle, with all the three sides as tangents, is called the incircle of the triangle. It is constructed as explained below.

Let us consider a scalene triangle PQR as shown. Draw the angle bisectors of all the internal angles of the triangle. As a property of the triangle, all of them are concurrent and the point of concurrence is called as incenter of the triangle and denoted as I. Since I happens to be a point on the angle bisectors of all the internal angles of the triangle, the perpendicular distances from I to all the sides of the triangle are congruent. In other words, a circle can be constructed inside the triangle, with all the three sides as tangents, by taking I as the center and IP (= IQ = IR = r) as the radius.

First let us study how a regular pentagon is constructed inscribing in a given circle. Refer to the following diagram.

Suppose a circle with center O as shown in the above diagram is the given circle. Let us consider the point A as one of the vertices of the pentagon to be constructed. Draw a ray AO and construct its perpendicular passing through O. Let the perpendicular intersect the circle at P and Q. Next, mark the mid point R of the line segment OQ.

Form R, draw an arc with radius equal to the measure RA. Let the arc intersect the line segment PQ at S.

Now, with A as center and and the measure AS as radius strike arcs which cut the circle at points B and E. With the same compass setting, draw arcs from B and E which cut the circle at C and D. Join, AB, BC, CD, DE and EA.

ABCDE is the required pentagon.

Draw a line segment AF of length equal to the given measure of the side of the hexagon.

Construct the perpendicular bisector of AF. With A as center and AF as radius, strike an arc which cuts the perpendicular bisector at O.

Now, with O as center and OA (= AF) as radius draw a complete circle.

From A and with the same radius strike an arc on the circle at B.

Repeat the same process to mark points C, D, and E.

Join AB, BC, CD, DE, EF and FA.

ABCDEF is the required hexagon.