- Home
- Class 10
- Class 10 Maths
- Triangles : Introduction : Theorems

Chapter Summary:

All congruent figures are similar, but it does not mean that all similar figures are congruent.

Two polygons of the same number of sides are similar, if:

- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio.

Two triangles are similar, if:

- Their corresponding angles are equal.
- Their corresponding sides are in the same ratio.

According to Greek mathematician Thales, "The ratio of any two corresponding sides in two equiangular triangles is always the same."

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.

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

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

According to the Indian mathematician Budhayan, "The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth)."

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

THEOREM 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.

Construction: ABC is a triangle. DE || BC and DE intersects AB at D and AC at E.

Join B to E and C to D. Draw DN ⊥ AB and EM ⊥ AC.

To prove:

Proof:

Area of triangle

Similarly,

Hence,

Similarly,

Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC.

Hence,

From above equations, it is clear that;

THEOREM 2:

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

Construction: ABC is a triangle in which line DE divides AB and AC in the same ratio. This means:

To prove: DE || BC

Let us assume that DE is not parallel to BC. Let us draw another line DE' which is parallel to BC.

Proof:

If DE' || BC, then we have;

According to the theorem;

Then according to the first theorem; E and E' must be coincident.

This proves: DE || BC

THEOREM 3:

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal. This means:

∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F

To prove:

Draw a line PQ in the second triangle so that DP = AB and PQ = AC

Proof:

Because corresponding sides of these two triangles are equal

This means; ∠ B = ∠ P = ∠ E and PQ || EF

This means;

Hence;

Hence,

THEOREM 4:

If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is also called SSS (Side-Side-Side) criterion.

Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means:

To prove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

And hence; Î” ABC ~ Î” DEF

In triangle DEF, draw a line PQ so that DP = AB and DQ = AC

Proof:

Because corresponding sides of these two triangles are equal

This means;

This also means; ∠ P = ∠ E and ∠ Q = ∠ F

We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q

Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F

From AAA criterion;

Î” ABC ~ Î” DEF proved

THEOREM 5:

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.

Construction: Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means;

∠ A = ∠ D and

To prove: Î” ABC ~ Î” DEF

Draw PQ in triangle DEF so that, AB = DP and AC = DF

Proof:

Because corresponding sides of these two triangles are equal

given∠ A = ∠ D

Hence; from SSS criterion

Hence;

Hence; Î” ABC ~ Î” DEF proved

THEOREM 6:

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Construction: Two triangles ABC and PQR are drawn so that, Î”ABC ~ Î” PQR.

To prove:

Draw AD ⊥ BC and PM ⊥ PR

Proof:

Hence;

Now, in Î” ABD and Î” PQM;

∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because Î” ABC ~ Î” PQR)

Hence; Î” ABD ~ Î” PQM

Hence;

Since, Î” ABC ~ Î” PQR

So,

Hence;

Similarly, following can be proven:

THEOREM 7:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Construction: Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:

Î” ABC ~ Î” ADB ~ Î” BDC

Proof:

In Î” ABC and Î” ADB;

∠ ABC = ∠ ADB

∠ BAC = ∠ DAB

∠ ACB = ∠ DBA

From AAA criterion; Î” ABC ~ Î” ADB

In Î” ABC and Î” BDC;

∠ ABC = ∠ BDC

∠ BAC = ∠ DBC

∠ ACB = ∠ DBC

From AAA criterion; Î” ABC ~ Î” BDC

Hence; Î” ABC ~ Î” ADB ~ Î” BDC proved.

THEOREM 8:

Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:

Proof:

In Î” ABC and Î” ADB;

Because these are similar triangles (as per previous theorem)

In Î” ABC and Î” BDC;

Adding equations (1) and (2), we get;

Proved.

- 10th Maths Paper Solutions Set 2 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Abroad Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE All India Previous Year 2015

Triangles will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

Introduction
>
NCERT Solution - Exercise 3.1
>
NCERT Solution - Exercise 3.2 Part-1
>
NCERT Solution - Exercise 3.2 Part-2
>
NCERT Solution - Exercise 3.3 Part-1
>
NCERT Solution - Exercise 3.3 Part-2
>
NCERT Solution - Exercise 3.4 Part-1
>
NCERT Solution - Exercise 3.4 Part-2
>
NCERT Solution - Exercise 3.5 Part-1
>
NCERT Solution - Exercise 3.5 Part-2
>
NCERT Solution - Exercise 3.6 Part-1
>
NCERT Solution - Exercise 3.6 Part-2
>

NCERT Solution - Exercise 4.1 Part-1
>
NCERT Solution - Exercise 4.1 Part-2
>
NCERT Solution - Exercise 4.2 Part-1
>
NCERT Solution - Exercise 4.2 Part-2
>
NCERT Solution - Exercise 4.2 Part-3
>
NCERT Solution - Exercise 4.3 Part-1
>
NCERT Solution - Exercise 4.3 Part-2
>
NCERT Solution - Exercise 4.3 Part-3
>
NCERT Solution - Exercise 4.3 Part-4
>
NCERT Solution - Exercise 4.3 Part-5
>
NCERT Solution - Exercise 4.4 Part-1
>
NCERT Solution - Exercise 4.4 Part-2
>

Solution of Exercise 5.1 (NCERT)-Part-1
>
Solution of Exercise 5.1 (NCERT)-Part-2
>
Solution of Exercise 5.1 (NCERT)-Part-3.1
>
Solution of Exercise 5.1 (NCERT)-Part-3.2
>
Solution of Exercise 5.2 (NCERT) Part -1
>
Solution of Exercise 5.2 (NCERT) Part -2
>
Solution of Exercise 5.2 (NCERT) Part -3
>
Solution of Exercise 5.2 (NCERT) Part -4
>
Solution of Exercise 5.2 (NCERT) Part -5
>
Solution of Exercise 5.2 (NCERT) Part -6
>
Solution of Exercise 5.3 (NCERT)Part -1
>
Solution of Exercise 5.3 (NCERT)Part - 2
>
Solution of Exercise 5.3 (NCERT)Part - 3
>
Solution of Exercise 5.3 (NCERT)Part - 4
>
Solution of Exercise 5.3 (NCERT)Part - 5
>
Solution of Exercise 5.3 (NCERT)Part - 6
>
Solution of Exercise 5.3 (NCERT)Part - 7
>
Solution of Exercise 5.3 (NCERT)Part - 8
>
Solution of NCERT Exercise 5.4 (Optional)
>

Introduction : Theorems
>
NCERT Solution of Exercise 6.1
>
NCERT Solution Exercise 6.2 Part - I
>
NCERT Solution Exercise 6.2 Part - II
>
NCERT Solution Exercise 6.2 Part - III
>
NCERT Solution of Exercise 6.3 Part - III
>
NCERT Solution of Exercise 6.4
>
NCERT Solution of Exercise 6.5
>
NCERT Solution of Exercise 6.5 Part - 2
>
NCERT Solution of Exercise 6.6 - Part - 1
>
NCERT Solution of Exercise 6.6 - Part - 2
>

THEOREM 1:
>
Solution of NCERT Exercise 10.1
>
Solution of NCERT Exercise 10.2
>
Solution of NCERT Exercise 10.2 - Part 2
>
Solution of NCERT Exercise 10.2 - part - 3
>
Solution of NCERT Exercise 10.2 - part - 4
>
Solution of NCERT Exercise 10.2 - part - 5
>
Solution of NCERT Exercise 10.2 - part - 6
>