# Similarity of Triangles

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The concept of similarity is particularly powerful with triangles. We already know that the corresponding internal angles in similar polygons are

For triangles, the **opposite is also true**: this means that **if you have two triangles with the same three angle sizes, then the triangles must be similar**.

Two triangles are similiar, if

**(i) their corresponding angles are equal and**

**(ii) their corresponding sides are in the same ratio (or proportion).**

The two triangles are said to be

The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the **Basic Proportionality Theorem** (now known as the Thales Theorem) for the same.

Let us perform the following activity to understand the Basic Proportionality Theorem:

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**Activity 2 :**

Any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB.

Now, through B, draw any line intersecting arm AY at C

Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from your constructions that

Measure AE and EC. Observe that

Thus, you can see that in △ABC, DE || BC and

Is it a coincidence?

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**Theorem 6.1:**

**If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.**

**Proof :**

We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively(seen in figure).

Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.Now, area of △ADE (1/2 x basexheight) =

x1 2 X area of △ADE is denoted as ar(ADE).So, ar(ADE) =

x AD X EN1 2 Similarly, ar(BDE) =

x1 2 X And ar(ADE) =

x1 2 X And ar(DEC) =

X1 2 X Therefore,

=ar ADE ar BDE Hence we get the answer is

and

ar ADE ar DCE Hence we get the answer is

Note that △BDE and DEC are on the same base DE and between the same parallels BC and DE. So, ar(BDE) = ar(DEC)

Therefore, from (1), (2) and (3), we have :

Is the converse of this theorem also true

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**Theorem 6.2 :**

**If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.**

This theorem can be proved by taking a line DE such that

If DE is not parallel to BC, draw a line DE' parallel to BC.

So,

Therefore

Adding 1 to both sides of above, you can see that E and E' must coincide.

If two angles in one triangle are congruent to two angles in another triangle, the two triangles are similar.

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**Example 1**

If a line intersects sides AB and AC of a ABC at D and E respectively and is parallel to BC, prove that

- Given DE || BC
- From (Theorem 6.1)
=AD DB / - Or,
=DB AD EC AE - add the value 1 for both sides
- Or,
- So

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**Example 2**

In Fig shown

- It is given that
- From (Theorem 6.2) ST ||
- Therefore, ∠PST = ∠
(Corresponding angles) - Also, it is given that
- So, ∠PRQ= ∠
[From (1) and (2)] - Therefore, PQ =
(Sides opposite the equal angles) - i.e., PQR is an isosceles triangle.

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**Example 3**

**ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AEED = BFFC**

- Let us join AC to intersect EF at G(shown in fig). AB ||
and EF || - So, EF ||
(Lines parallel to the same line are parallel to each other) - Now, in △ADC, EG ||
(As EF || DC) - So,
=AE ED / (Theorem 6.1) - Similarly, from △ CAB, =
- Therefore
=AG GC / - Therefore, from (1) and (2),
=AE ED /