CBSE Class 10 Mathematics chapter 6 - Triangle - Similarity of Triangles

 Similarity of Triangles:

Two triangles are similiar, if

1. their corresponding angles are equal and
2. their corresponding sides are in the same ratio (or proportion).

Thales Theorem or Basic Proportionality Theorem:

If a line is drawn parallel to one side of a triangle and it intersects the other two sides, it divides those sides proportionally.

Let, if DE is parallel to BC in ΔABC and it intersect AB and AC at point D and E, then

Construction:

(i) Join the vertex B of ΔABC to D and the vertex C to E to form the lines BE and CD.

(ii) Draw a perpendicular EN to the side AB and DM⊥AC as per given figure.

Proof:

We know that, Area of a triangle= ½ × Base × Height

So,

The area of ∆ADE = ½ × AD × EN

Similarly, area of ∆DBE= ½ × DB × EN

Area of ∆ADE = ½ × AE × DM

Also, area of ∆ECD = ½ × EC × DM

Now, if we find the ratio of the area of triangles ∆ADEand ∆DBE, we have

As per the property of triangles “the triangles drawn between the same parallel lines and on the same base have equal areas”.

So, we can say that ∆DBE and ∆ECD have the same area.

Area of ∆DBE = area of ∆ECD …………..(iii)

Therefore, from the equations (i), (ii) and (ii) we can say that,

The Mid Point theorem:

• It is also known as the Mid-segment Theorem.
• When in a triangle, the line segment joining the midpoints of two sides is parallel to the third side, and its length is half the length of the third side.
• If M is the midpoint of AB and N is the midpoint of AC in triangle ABC, then,

MN ∥ BC and MN = ½BC

Proof of Parallelism:

• Since M and N are midpoints, we know that AM = MB and AN = NC.
• Therefore, by the Mid-segment Theorem, MN is parallel to BC.

Proof of Length:

• Construct AD as the median from A to BC, intersecting BC at D.
• By the Midpoint Definition: AM = MB and AN = NC because M and N are midpoints.
• BD = CD (medians in a triangle are equal).
• MN is parallel to BC by the Midsegment Theorem.
• In triangle MBC, MN is parallel to BC and MB = MD, so MN = ½ BC by the Converse of the Triangle Proportionality Theorem.

Converse of Basic Proportionality Theorem: As per this theorem If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. Let, A triangle ABC with a line DE intersecting side AB and AC at D and E, respectively, Assume for the sake of contradiction that DE is not parallel to BC. This means EC = E’C, which is possible only when E and E’ must coincide. However, we assumed DE′ is parallel to BC and E and E′ coinciding would mean DE is also parallel to BC. This contradiction implies that our assumption was incorrect. Therefore, DE must be parallel to BC. Hence, the statement is proved.

Example 1.	If a line intersects sides AB and AC of a ΔABC at D and E respectively and is parallel to BC, prove that  $$\frac{AD}{AB}=\frac{AE}{AC}$$
Solution:	Given DE ∥ BC $$\frac{AD}{DB}=\frac{AE}{CE} ...(by Thales Theorem)$$ $$Or, \frac{DB}{AD}=\frac{EC}{AE}$$ By adding 1 both side $$\Rightarrow \frac{DB}{AD}+1=\frac{EC}{AE}+1$$ $$\Rightarrow\frac{DB+AD}{AD}=\frac{EC+AE}{AE}$$ $$\Rightarrow\frac{AB}{AD}=\frac{AC}{AE}$$ $$So,\Rightarrow\frac{AD}{AB}=\frac{AE}{AC}$$ Hence, proved.

Example 2.

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  $$\frac{AE}{ED}=\frac{BF}{FC}$$

Solution:

Given AB || DC and EF || AB Let draw AC to intersect EF at point G as per figure. We know that Lines parallel to the same line are parallel to each other So, EF || DC In ΔADC because of EF || DC so EG || DC Therefore, according to Thales theorem, we can write $$So,\frac{AE}{ED}=\frac{AG}{GC}..............(i)$$ Similarly in ΔCAB $$\frac{CG}{AG}=\frac{CF}{BF}$$ $$\frac{AG}{GC}=\frac{BF}{FC}...............(ii)$$ From equation (i) and (ii)  $$\frac{AE}{ED}=\frac{BF}{FC}$$

Example 3.

In given figure,  $$\frac{PS}{SQ}=\frac{PT}{TR}$$ and $$\angle PST=\angle PRQ.$$ Prove that PQR is an isosceles triangle.

Solution:

Given: $$\frac{PS}{SQ}=\frac{PT}{TR}$$ We know that If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. So, ST || QR because ∠PST and ∠PQR are corresponding angle. So, $$\angle PST=\angle PQR ........................(i)$$ It is given that $$\angle PST=\angle PRQ.........................(ii)$$ From (i) and (ii) $$\angle PRQ=\angle PQR$$ Therefore, PQ and PR are sides of opposite equal angles. So, PQ = PR Hence, ΔPQR is an isosceles triangle.