Geometry is a skill of the eyes, hands and that of the mind. Any construction in geometry involves drawing a diagram based on provided data. In this module, we will learn a few skills regarding construction related to the line segment and triangles. We will also revise a few properties of geometry while doing construction.

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## Construction Related to Line Segments

You must have seen many times that a point divides a line segment in a particular ratio. In this module, we will learn how to divide a given line segment in the desired ratio x:y without using a ruler.

### Dividing a given line segment in a particular ratio

Let’s say we have been given a line segment AB which we want to divide it by a ratio of 3:2. Here, x = 3 and y = 2. To construct such a line segment, follow the steps as given below:

### Steps of Construction

- Draw any ray AZ making an acute angle with AB.
- Locate 5 points (
**x + y = 3 + 2 = 5**), say, A_{1}, A_{2}, A_{3}, A_{4}, and A_{5}using a compass on AZ such that these points are equally distanced from each other.- Open the compass of a specific width and make an arc on the ray AZ keeping the needle on the point A. The point where the arc intersects with AZ is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the steps 3 more times making AA
_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}.

- Open the compass of a specific width and make an arc on the ray AZ keeping the needle on the point A. The point where the arc intersects with AZ is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the steps 3 more times making AA
- Join A
_{5}with B making a line segment A_{5}B. (See Figure 1) - Since we need to divide the line segment in 3:2, draw a line through the point A
_{3}which is parallel to A_{5}B by making ∠AA_{3}C = ∠AA_{5}B such that A_{3}C intersects AB at the point C.- To construct a parallel line using
**Angle Copy Method**, adjust the width of the compass to roughly half of the length of A_{4}A_{5}. Keeping the needle at point A_{5}, make an arc cutting the ray AZ and the line segment A_{5}B. This is arc 1. - Make a similar arc from point A
_{3}using the same compass width. This is arc 2. - Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AZ, the pencil should be on the point where arc 1 meets the line segment A
_{5}B. - Keeping the same width, cut the arc 2 making an intersecting point G.
- Make a line segment from A
_{3}through G meeting the line segment AB at C. The line segment A_{3}C || A_{5}B.

- To construct a parallel line using
- Therefore, AC : CB = 3 : 2. (See Figure 1)

**Rationale and Proof**

Since A_{3}C || A_{5}B

∴ \( \frac{AA_3}{A_3A_5} \) = \( \frac{AC}{CB} \) [* Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally*]

\( \frac{AA_3}{A_3A_5} \) = \( \frac{3}{2} \) *[By construction]*

∴ \( \frac{AC}{CB} \) = \( \frac{3}{2} \)

## Alternate Method

Following is the alternative method to divide a given line segment AB into a said ratio (same as above, that is, 3:2).

**Steps of Construction**

- Draw any ray AG making an acute angle with AB.
- Draw a ray BH parallel to AG by making ∠ABH equal to ∠BAG using the same method as described above in steps from 4(a) to 4(e).
- Locate the pointsA
_{1}, A_{2}, and A_{3}**(x = 3)**on AG and B1, B2**(y = 2)**on BH such that AA_{1}= A_{1}A_{2}= A_{2}A_{3}= BB_{1}= B_{1}B_{2}using the same step as described in 2(a) above. - Join A
_{3}B_{2}. Let it intersect AB at a point C (See Figure 2). - Then AC : CB = 3 : 2

**Rationale and Proof**

Here,

∠CAA_{3} = ∠CBB_{2} [*By construction*]

∠ACA_{3} = ∠BCB_{2} [*Vertically opposite angles are always equal*]

∠AA_{3}C = ∠BB_{2}C [A*lternate interior angles within two parallel lines are always equal*]

∴ ∆ACA_{3} ~ ∆BCB_{2} [*By AAA criteria of triangle similarity*]

⇒ \( \frac{AC}{BC} \) = \( \frac{CA_3}{CB_2} \) = \( \frac{AA_3}{BB_2} \)

Now, using construction \( \frac{AA_3}{BB_2} \) = \( \frac{3}{2} \)

∴ \( \frac{AC}{BC} \) = \( \frac{3}{2} \)

**Construction Related to Triangles**

We have studied about similar triangles and their properties in previous classes. Have you ever wondered how these similar triangles are designed? Let’s take a look how.

### Constructing a triangle similar to a given triangle

There are a few ways by which we can draw a triangle similar to the given one. Here, we shall study how to construct the same using the** Scale Factor method** without the help of a ruler.

Scale Factor is the ratio of two corresponding sides of the similar triangles. For example, let’s say that ∆ABC is similar to ∆DEF (∆ABC ~ ∆DEF) wherein AB = 3, BC = 4, and AC = 5; whereas DE = 6, EF = 8 and DF = 10. Here, the Scale Factor of two corresponding sides is

\( \frac{AB}{DE} \) = \( \frac{BC}{EF} \) = \( \frac{AC}{DF} \) = \( \frac{1}{2} \)

So, let’s see how we can construct a triangle similar to the given triangle (∆PQR) with a scale factor of \( \frac{3}{4}\), that is, to construct a smaller triangle (∆P′QR′) compared to ∆PQR wherein the length of the sides of ∆P′QR′ will be 3/4^{th} to that of ∆PQR.

### Steps of Construction using Scale Factor

- Draw any ray QX making an acute angle with QR on the side opposite to the vertex P.
- Locate 4 points Q
_{1}, Q_{2}, Q_{3}, and Q_{4}on QX so that QQ_{1}= Q_{1}Q_{2}= Q_{2}Q_{3}= Q_{3}Q_{4}using the same method as described above in step 2(a) (Dividing a line segment in a given ratio). Note: We located 4 points because 4 is the greater value in the Scale Factor ratio of \( \frac{3}{4} \) - Considering the given triangle, ∆PQR, join a line Q
_{4}R. - Draw a line through Q3 parallel to Q
_{4}R using the same method as described above in steps from 4(a) to 4(e) (Dividing a line segment in a given ratio) intersecting the line QR at R′. This gives us Q_{3}R′ || Q_{4}R and ∠QQ_{3}R′ = ∠QQ_{4}R (See Figure 3). - Similarly, draw a line through R′ parallel to the line PR intersecting the line PQ at a point, P′. This gives us P′R′ || PR and ∠P′R′Q = ∠PRQ (See Figure 3).
- Subsequently, ∆P′QR′ is the required triangle similar to ∆PQR with Scale Factor of \( \frac{3}{4} \).

**Rationale and Proof**

Since Q_{3}R′ || Q_{4}R and QQ_{3} is proportionate to QQ_{4} in the ratio of 3:4 (*by construction*)

∴ \( \frac{QQ_3}{QQ_4} \) = \( \frac{QR’}{QR} \) = \( \frac{3}{4} \) [*By Triangle Proportionality Theorem*]

Here,

∠P′QR′ = ∠PQR [*Common angle]*

∠P′R′Q = ∠PRQ [*By Construction*]

∠R′P′Q = ∠RPQ [*If two corresponding angles of two triangles are equal, the third angle will always be equal*]

∴ ∆P′QR′ ~ ∆PQR [*By AAA criteria of triangle similarity*]

Since ∆P′QR′ ~ ∆PQR and \( \frac{QR’}{QR} \) = \( \frac{3}{4} \)

∴ ∆P′QR′ ~ ∆PQR with Scale Factor of \( \frac{3}{4} \).

## Solved Examples for You

**Question 1: Given a line segment AB. Find a point, describing the steps, which divides the line segment in 2:1 without using a ruler.**

**Answer :** Steps of Construction

- Draw any ray AX, making an acute angle with AB.
- Locate 3 points (
**x + y = 2 + 1 = 3**), say, A_{1}, A_{2}, and A_{3 }using a compass on AX such that these points are equally distanced from each other.- Open the compass of a specific width and make an arc on the ray AX keeping the needle on the point A. The point where the arc intersects with AX is now A
_{1}. Repeat the step keeping the same width from A_{1}making a point A_{2}. Repeat the step one more times making AA_{1}= A_{1}A_{2}= A_{2}A_{3}.

- Open the compass of a specific width and make an arc on the ray AX keeping the needle on the point A. The point where the arc intersects with AX is now A
- Join A
_{3}with B making a line segment A_{3}B. (See Figure A) - Since we need to divide the line segment in 2:1, draw a line through the point A
_{2}which is parallel to A_{3}by making ∠AA_{2}B = ∠AA_{3}B intersecting AB at the point C.- To construct a parallel line, adjust the width of the compass to roughly half of the length of A
_{2}A_{3}. Keeping the needle at point A_{3}, make an arc cutting the ray AX and the line segment A_{3}B. This is arc 1. - Make a similar arc from point A
_{2}using the same compass width. This is arc 2. - Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AX, the pencil should be on the point where arc 1 meets the line segment A
_{3}B. - Keeping the same width, cut the arc 2 making an intersecting point O.
- Make a line segment from A
_{2}through O meeting the line segment AB at C. The line segment A_{2}C || A_{3}B.

- To construct a parallel line, adjust the width of the compass to roughly half of the length of A
- This point C divides AB as AC : CB = 2 : 1. (See Figure A)

**Question 2: What do you mean by segment in maths?**

**Answer:** Segment refers to the part of a line that connects two points. It happens to the shortest distance that exists between the two points. Furthermore, segment has a length. The word “segment” is important because a line can go on continuously without end. In contrast, a line segment has definite endpoints.

**Question 3: Explain what is meant by segments of circle?**

**Answer:** A chord of a circle divides a circle into two parts. These parts are known as the segments of circle.

**Question 4: How can one measure a line segment?**

**Answer:** A line segment is nothing more than a part of a line. In order to measure the length of a line segment, put the segment’s endpoint on the ruler’s zero mark. They must see where it ends. In inches, one must round to the nearest ¼, while in centimeters; one must measure to the nearest centimeter.

**Question 5: How can one find out the midpoint of a segment?**

**Answer:** One can find out the midpoint of a segment by dividing the segment’s length by 2 and counting that value from any endpoint.