## Construction Of Similar Triangle As Per Given Scale Factor

Scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.
This construction involves two different situations :
(i) The triangle to be constructed is smaller than the given triangle, here scale factor is less than 1.
(ii) The triangle to be constructed is bigger than the given triangle, here scale factor is greater than 1.

## Construct A Triangle Similar To A Given Triangle Examples

Example 1:   Construct a ∆ABC in which AB = 4 cm, BC = 5 cm and AC = 6 cm. Now, construct a triangle similar to ∆ABC such that each of its sides is two-third of the corresponding sides of ∆ABC. Also, prove your assertion.
Solution.    Steps of construction
Step I: Draw a line segment AB = 4 cm.
Step II: With A as centre and radius = AC = 6 cm, draw an arc.
Step III: With B as centre and radius = BC = 5 cm, draw another arc, intersecting the arc drawn in step II at C.
Step IV: Join AC and BC to obtain ∆ABC.
Step V: Below AB, make an acute angle ∆BAX.
Step VI: Along AX, mark off three points (greater of 2 and 3 in 2/3) A1, A2, A3  such that AA1 = A1A2 = A2A3.
Step VII: Join A3B.
Step VIII: Since we have to construct a triangle each of whose sides is two-third of the corresponding sides of ∆ABC. So, take two parts out of three equal parts on AX i.e. from point A2, draw
A2B’ || A3B, meeting AB at B’.
Step IX: From B’, draw B’C’ || BC, meeting AC at C’.
AB’C’ is the required triangle, each of the whose sides is two-third of the corresponding sides of ∆ABC.
Justification: Since B’C’ || BC.
So, ∆ABC ~ ∆AB’C’
$$\frac{B’C’}{BC}=\frac{AC’}{AC}=\frac{AB’}{AB}=\frac{2}{3}\text{ }\left[ \frac{AB’}{AB}=\frac{2}{3} \right]$$
Let ABC be the given triangle and we want to construct a triangle similar to ∆ABC such that each of its sides is $$\frac { m }{ n }$$ th of the corresponding sides of ∆ABC such that m < n. We follow the following steps to construct the same.
Steps of construction when m > n.

Step I: construct the given triangle by using the given data.
Step II: Take any of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle.
Step III: At one end, say A, of base AB construct an acute angle ∠BAX below base AB i.e. on the opposite side of the vertex C.
Step IV : Along AX, mark-off m (large of m and n) points A1, A2,…,Am on AX such that AA1 = A1A2 = ….. = Am-1Am.
Step V : Join An to B and draw a line through Am Parallel to AnB, intersecting the extended line segment AB at B’.
Step VI: Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’.
Step VII: ∆AB’C’ so obtained is the required triangle.
Justification: For justification of the above construction consider triangles ABC and AB’C’. In these two triangles, we have
∠BAC = ∠B’AC’
∠ABC = ∠AB’C’           [∵ B’C’ || BC]
So, by AA similarity criterion, we have
∆ABC ~ ∆AB’C’
$$\Rightarrow \frac{AB}{AB’}=\frac{BC}{B’C’}=\frac{AC}{AC’}\text{ }………\text{ (i)}$$
In D A AmB’,  AnB || AmB’.
$$\frac{AB}{BB’}=\frac{A{{A}_{n}}}{{{A}_{n}}{{A}_{m}}}$$
$$\Rightarrow \frac{BB’}{AB}=\frac{{{A}_{n}}{{A}_{m}}}{A{{A}_{n}}}\text{ }\Rightarrow \text{ }\frac{BB’}{AB}=\frac{m-n}{n}$$
$$\Rightarrow \frac{AB’-AB}{AB}=\frac{m-n}{n}\text{ }\Rightarrow \text{ }\frac{AB’}{AB}-1=\frac{m-n}{n}$$
$$\Rightarrow \frac{AB’}{AB}=\frac{m}{n}\text{ }……..\text{ (ii)}$$
From (i) and (ii), we have
$$\frac{AB’}{AB}=\frac{B’C’}{BC}=\frac{AC’}{AC}=\frac{m}{n}$$

Example 2:    Draw a triangle ABC with side BC = 7 cm,
∠B = 45º, ∠A = 105º. Then construct a triangle whose sides are (4/3) times the corresponding sides of ∆ABC.
Solution.    In order to construct ∆ABC, we follow the following steps:

Step I: Draw BC = 7 cm.
Step II: At B construct ∠CBX = 45º and at C
construct ∠BCY = 180º – (45º – 105º) = 30º
Suppose BX and CY intersect at A. ∆ABC so obtained is the given triangle. To construct a triangle similar to ∆ABC,
we follow the following steps.
Step I: Construct an acute angle ∠CBZ at B on opposite side of vertex A of ∆ABC.
Step II: Mark-off four (greater 4 and 3 in 4/3) points
B1, B2, B3, B4 on BZ such that
BB1 = B1B2 = B2B3 = B3B4.
Step III: Join B3 (the third point) to C and draw a line through B4 parallel to B3C, intersecting the extended line segment BC at C’.
Step IV: Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.
Triangle A’BC’ so obtained is the required triangle such that
$$\frac{A’B}{AB}=\frac{BC’}{BC}=\frac{A’C’}{AC}=\frac{4}{3}$$

Example 3:    Construct a triangle similar to a given triangle ABC such that each of its sides is (6/7)th of the corresponding sides of ∆ABC. It is given that
AB = 5 cm, AC = 6 cm and BC = 7 cm.
Solution.    Steps of Construction

Step I: Draw a line segment BC = 7 cm.
Step II: With B as centre and radius = AB = 5 cm, draw an arc.
Step III : With C as centre and radius = AC = 6 cm, draw another arc, intersecting the arc drawn in step II at A.
Step IV: Join AB and AC to obtain the triangle ABC.
Step V: Below base BC, construct an acute angle ∠CBX.
Step VI: Along BX, mark off seven points B1, B2, B3, B4, B5, B6, B7 such that
BB1 = B1B2 = …… = B6B7.
Step VII: Join B7C.
Step VIII: Since we have to construct a triangle each of whose sides is (6/7)th of the corresponding sides of ∆ABC. So take 6 parts out of 7 equal parts on BX i.e. from B6, Draw B6C´ || B7C, intersecting BC at C´.
Step IX: From C´, draw C´A´ || CA, meeting BA at A´.
∆A´BC´ is the required triangle each of whose sides is (6/7)th of the corresponding sides of ∆ABC.

Example 4:    Construct a ∆ABC in which AB = 4 cm, ∠B = 60º and altitude CL = 3 cm. Construct a ∆ADE similar to ∆ABC such that each side of ∆ADE is 3/2 times that of the corresponding side of ∆ABC.
Solution.    Steps of construction

Step I: Draw a line segment AB = 4 cm.
Step II: Construct ∠ABP = 60º.
Step III: Draw a line GH || AB at a distance of 3 cm, intersecting BP at C.
Step IV: Join CA.
Thus, ∆ABC is obtained.
Step V: Extend AB to D such that AD = 3/2 AB = $$\left( \frac{3}{2}\times 4 \right)$$ cm = 6 cm.
Step VI: Draw DE || BC, cutting AC produced at E.
Then ∆ADE is the required triangle similar to ∆ABC such that each side of ∆ADE is 3/2 times the corresponding side of ∆ABC.
Proof: Since DE || BC,
we have ∆ADE ~ ∆ABC.
$$\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}=\frac{3}{2}$$

## Strategies for Dealing with Similar Triangles

Triangles are similar if their corresponding (matching) angles are congruent (equal in measure) and the ratio of their corresponding sides are in proportion.

There are many different types of problems that involve similar triangles. And, fortunately, there are several different ways to arrive at an answer.
Let’s look at some strategies for arriving at answers!

Style 1: The similar triangles are two separate triangles:
Find x:

Create a proportion matching the corresponding sides.

1. Small triangle on top:
$$\frac { 10 }{ x } =\frac { 6 }{ 12 }$$
x = 20
2. Large triangle on top:
$$\frac { x }{ 10 } =\frac { 12 }{ 6 }$$
x = 20

HINT: These two triangles are sitting such that their corresponding parts are in the same position in each triangle. If the triangles are not sitting in this manner, you can match the corresponding sides by looking across from the angles which are equal in each triangle.

Style 2: The similar triangles overlap:

Many problems involving similar triangles have one triangle ON TOP OF (overlapping) another triangle.
Since $$\overline { DE }$$ is marked to be parallel to $$\overline { AC}$$, we know that we have ∠BDE congruent to ∠DAC (by corresponding angles). ∠B is shared by both triangles, so the two triangles are similar by AA.
There are two ways to attack this type of problem.
Easiest method to use:
Use FULL sides of the two triangles when dealing with the problem. Do not use DA or EC since they are not sides of triangles.
Easy to forget:
Use a theorem relating to parallel lines, which says that If a line is parallel to one side of a triangle, and intersects the other two sides, the line divides these two sides proportionally.

Let’s try some problems with this type of diagram:

Find x.
This problem MUST use the full sides of triangles as a solution. The parallel theorem does not work here. The problem asks you to find x where x is a FULL side.
Here is the solution:

$$\frac { 5 }{ 15 } =\frac { x }{ 15 }$$
x = 5

## Similar Triangles

Objects, such as these two cats, that have the same shape, but do not necessarily have the same size, are said to be “similar“.

The mathematical symbol used to denote similar is ~.
Definition: In mathematics, polygons are similar if their corresponding (matching) angles are congruent (equal in measure) and the ratio of their corresponding sides are in proportion.
(This definition allows for congruent figures to also be “similar“, where the ratio of the corresponding sides is 1:1.)

## Basic Proportionality Theorem or Thales Theorem

Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Given: A triangle ABC in which DE || BC, and intersects AB in D and AC in E.

## Converse of Basic Proportionality Theorem

Statement: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Given: A DABC and a line l intersecting AB in D and AC in E,

## Basic Proportionality Theorem Example Problems With Solutions

Example 1:    D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC.
Find the value of x, when
(i) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm
(ii) AD = (7x – 4) cm, AE = (5x – 2) cm,
DB = (3x + 4) cm and EC = 3x cm.
Solution:

Example 2:    Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB × TC.
Solution:    In ΔTXM, we have

Example 3:    In fig., EF || AB || DC. Prove that $$\frac{AE}{ED}=\frac{BF}{FC}$$.
Solution:    We have, EF || AB || DC

Example 4:    In figure, ∠A = ∠B and DE || BC. Prove that AD = BE
Solution:

Example 5:    In fig., DE || BC. If AD = 4x – 3, DB = 3x – 1, AE = 8x – 7 and EC = 5x – 3, find the value of x.
Solution:

Example 6:    Prove that the line segment joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:
Given : A quadrilateral ABCD in which P, Q, R, S are the midpoints of AB, BC, CD and DA respectively.
To prove: PQRS is a parallelogram.

Example 7:    In fig. DE || BC and CD || EF. Prove that AD2 = AB × AF.
Solution:

Example 8:    Ex.8 In the given figure PA, QB and RC each is perpendicular to AC such that PA = x,
RC = y, QB = z, AB = a and BC = b. Prove that $$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$.
Solution:

Example 9:    In fig., LM || AB. If AL = x – 3, AC = 2x, BM = x – 2 and BC = 2x + 3, find the value of x.
Solution:

Example 10:    In a given ∆ABC, DE || BC and $$\frac{AD}{DB}=\frac{3}{4}$$. If AC = 14 cm, find AE.
Solution:

Example 11:    In figure, DE || BC. Find AE.
Solution:

Example 12:    In figure, ABC is a triangle in which AB = AC. Points D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic.
Solution:    In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that

Example 13:    In fig., $$\frac{AD}{DB}=\frac{1}{3}\text{ and }\frac{AE}{AC}=\frac{1}{4}$$. Using converse of basic proportionality theorem, prove that DE || BC.
Solution:

Example 14:    Using basic proportionality theorem, prove that the lines drawn through the points of trisection of one side of a triangle parallel to another side trisect the third side.
Solution:

Example 15:    In the given figure, $$\frac{AD}{DB}=\frac{AE}{EC}$$ and ∠ADE = ∠ACB. Prove that ∆ABC is an isosceles triangle.
Solution:

Example 16:    In fig., if DE || AQ and DF || AR. Prove that EF || QR.
Solution:

Example 17:    Two triangles ABC and DBC lie on the same side of the base BC. From a point P on BC, PQ || AB and PR || BD are drawn. They meet AC in Q and DC in R respectively. Prove that QR || AD.
Solution:    Given: Two triangles ABC and DBC lie on the same side of the base BC. Points P, Q and R are points on BC, AC and CD respectively such that PR || BD and PQ || AB.

Example 18:    ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that $$\frac{AE}{ED}=\frac{BF}{FC}$$
Solution:     Given: A trap ABCD in which AB || DC.
E and F are points on AD and BC respectively such that EF || AB.

Example 19:    In fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:

Example 20:    Any point X inside ∆DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF.
Solution:    A ΔDEF and a point X inside it. Point X is joined to the vertices D, E and F. P is any point on DX. PQ || DE and QR || EF.

Thus, in ΔXFD, points R and P are dividing sides XF and XD in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have, PR || DF

Example 21:    Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
Solution:    Given: A trapezium ABCD in which DC || AB and EF is a line parallel to DC and AB.

Example 22:    Prove that the line drawn from the mid-point of one side of a triangle parallel of another side bisects the third side.
Solution:    Given: A DABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.
To Prove : E is the mid-point of AC i.e.,
AE = EC.

Hence, E bisects AC.

Example 23:    Prove that the line joining the mid-point of two sides of a triangle is parallel to the third side.
Solution:    Given: A ΔABC in which D and E are mid-point of sides AB and AC respectively.

Thus, the line DE divides the sides AB and AC of ΔABC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have
DE || BC

Example 24:    AD is a median of ∆ABC. The bisector of ∠ADB and ∠ADC meet AB and AC in E and F respectively. Prove that EF || BC.
Solution:    Given: In ∆ABC, AD is the median and DE and DF are the bisectors of ∠ADB and ∠ADC respectively, meeting AB and AC in E and F respectively.
To Prove: EF || BC
Proof: In ∆ADB, DE is the bisector of ∠ADB.

Thus, in ∆ABC, line segment EF divides the sides AB and AC in the same ratio.
Hence, EF is parallel to BC.

Example 25:    O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA.
Solution: