# Triangle class 10 theorems[With Step-by-Step Proof]

In earlier classes, we have read about the congruent triangle in detail. We learned about congruency criteria for the triangle like AAA, SSS, ASA, etc. Now, in class 10th we will study similar triangles. In this article, you will learn about triangle class 10 theorems and see their proof

Q.1 What is the difference between congruent and similar triangles?

Ans.

Congruent triangles: When two triangles have the same shape and size. Then, triangles are called congruent triangles.

Similar triangles: When two triangles have the same shape but different sizes. Then, Triangles are called similar triangles.

For Example: More Examples…….

(i) Any two squares are always similar(because they have the same shape)but need not be congruent(Because To be congruent, the shape and size of both Square should be the Same). (ii) Any two circles are similar but not necessarily congruent. If their radii are equal, then both circles will congruent. Note: All congruent figures are similar but all similar figures need not be congruent.

### Similar polygon

If you have forgotten the basic definition of Polygon, No problem! I explain

Polygon: A plane figure with three or more sides, angles is called a polygon.

Example: Triangle, square, pentagon, hexagon, rectangle, rhombus, etc are Polygon.

Criteria for Similarity

Two polygons are said to be similar, if

(i) Their corresponding angles are equal, and

(ii) The length of their corresponding sides are proportional(having the same ratio)

Suppose, I have two triangles ΔABC and ΔDEF. For similarity

(i) ∠A=∠D,∠B=∠E, and ∠C=∠F

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

Note: If the corresponding angles of two triangles are equal, then they are known as equiangular triangles.

• All Equilateral triangles are Equiangular Triangles

### Triangle class 10 theorems

Here are some triangle class 10 theorems that you must practice

Theorem 6.1 :(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.

Proof:

Given: A triangle ABC in which DE intersects AB in D and AC in E. Construction: Draw EM⊥BA, DN⊥AC, and join BE, CD. Proof:

Since EM is perpendicular to AB. Therefore, EM is the height of ΔADE and ΔDBE

We know the formula for the area of a right-angled triangle

Area of right-angled triangle=1/2×base×height

So

and,

Area of triangleΔDBE=1/2×base×height

Area of triangleΔDBE=1/2×BD×EM

Similarly

and,

Area of a triangle(ΔDEC)=1/2×EC×DN

Therefore

But, ΔDBE and ΔDEC are on the same base DE and between the same parallel DE and BC You have already read a theorem in class 9 If Two triangles are on the same base and between the same parallel line. then, the area of both triangles will be equal.

∴         Area(ΔDBE)=Area(ΔDEC)

Subtitute Area(ΔDBE)=Area(ΔDEC) in equation(i)

As you see LHS of both equation are equal .So,RHS will also equal

### Theorem 6.2:(converse of basic proportionality theorem or Thales 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 line

Given: A ΔABC in which a line divides both side of triangle in same ratio  \frac{AD}{BD} =\frac{AE}{EC} To prove: DE\parallel BC

Proof: We will prove this theorem with the help of a contradiction method.

So let DE is not parallel to BC. Then, there must be another line that is parallel to BC. Let DF\parallel BC

Since DF\parallel BC.So, by Thales theorem(Proved Above)

But it is given,

Again LHS of both equation are equal

\frac{AF}{FC}=\frac{AE}{EC}

Now ,Add 1 on both sides

⇒  \frac{AF}{FC}+1=\frac{AE}{EC} +1          (Adding 1 on both sides)

\frac{AF+FC}{FC} =\frac{AE+EC}{EC}     (See figure)

\frac{AC}{FC} =\frac{AC}{EC}

FC=EC

This is only possible when F and E coincide.

So, our assumption is wrong

Hence  DE\parallel BC

### Theorem. (Angle bisector theorem)

Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle

Given: A ΔABC in which AD is the internal bisector of ∠A and meet BC at D To prove: This bisector divides opposite sides in the ratio of the sides containing the angle.

Construction: Draw CE parallel to DA to meet BA produced in E Proof: Since CE parallel to DA and AC cuts them

∠2=∠3              …(i)   {Alternate angle }

∠1=∠4              …(ii)  {Corresponding angle}

It is given,

∠1=∠2                           {AD is the bisector of ∠A}

From equation (i) and (ii) ,we get

∠3=∠4

Therefore            AE=AC             ………..(iii)      [Sides opposite to equal angles are equal]

Since in ΔBCE ,We have DA || CE

\frac{BD}{DC} =\frac{BA}{AE}      [Using Thales theorem]

\frac{BD}{DC} =\frac{AB}{AC}      [From Eq(iii)]

Hence ,  \frac{BD}{DC} =\frac{AB}{AC}

### Theorem  :(Converse of angle bisector theorem)

Statement: In a triangle ABC ,if D is a point on BC such that \frac{BD}{DC} =\frac{AB}{AC},prove that AD is the bisector of ∠A.

Given: A ΔABC ,in which D is a point on BC such that \frac{BD}{DC} =\frac{AB}{AC} To prove: AD is the bisector of ∠A

Construction: Produce BA to E such that AE=AC.Join EC Proof: In ΔACE, we have

AE=AC                                                   [By construction]

Therefore  ∠3=∠4

It is given

\frac{BD}{DC} =\frac{AB}{AC}

It becomes

\frac{BD}{DC} =\frac{AB}{AE}      ( AE=AC)

Thus, in ΔBCE, we have

\frac{BD}{DC} =\frac{AB}{AE}

Therefore, by the converse of Thales’s theorem

DA parallel to CE

∠1=∠4                       …….(ii)     (corresponding angle)

∠2=∠3                       ………(iii)    ( Alternate angle)

But ,   ∠3=∠4                {From (i)}

Therefore from equation(i),(ii)and (iii)

∠1=∠2

Hence, AD is the bisector of ∠A.

### Theorem :(Mid-point theorem)

Statement: The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.

Given: A ΔABC in which D is the mid-point of side AB and the line DE is drawn parallel to BC meting AC in E. To prove: E is the mid-point of AC i.e AE=EC

Proof: In ΔABC, we have

DE\parallel BC

But,   D is the mid-point of AB

From equation(i) and (ii) ,we get

\frac{AE}{EC}=1

∴              AE=EC

Hence ,E bisects AC

### Theorem 6 :(Converse of mid-point theorem)

Statement: The line joining the mid-points of two sides of a triangle is parallel to the third side.

Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively To prove: DE\parallel BC

Proof: Since D and E are mid-points of 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 Thales’s theorem, we obtain DE||BC

## Triangle class 10 similarity theorems

These are 3 important theorems of similarity for class 10 . See their statements and proofs

Theorem.6.3 (AAA similarity criteria)

Statement: If in two triangles ,the corresponding angles are equal ,then their corresponding sides are proportional and hence the triangles are similar

Given: Δ ABC and DEF such that ∠A=∠D ,∠B=∠E and ∠C=∠F

To prove: ΔABC∼ΔDEF

Construction: Cut DP=AB and DQ=AC .Join PQ Proof:

In ΔABC and ΔDPQ ,we have

AB=DP  [By construction]

AC=DQ [By construction]

∠A = ∠D

∴ ΔABC ≅ ΔDPQ  [By SAS congruency criteria]

⇒∠B=∠P [By C.P.C.T]

⇒∠E=∠P [Given ∠B=∠E]

Therefore , PQ||EF      [Corresponding angles are equal ]

\frac{DP}{DE} =\frac{DQ}{DF}

\frac{AB}{DE} =\frac{CA}{FD}   [DP=AB and DQ=AC]

Similarly we can prove

\frac{AB}{DE} =\frac{BC}{EF}

\frac{AB}{DE} =\frac{BC}{EF}=\frac{CA}{FD}

Thus , ∠A=∠D ,∠B=∠E , ∠C=∠F and \frac{AB}{DE} =\frac{BC}{EF}=\frac{CA}{FD}

Hence ΔABC ∼ ΔDEF

Theorem 6.4  : (SSS similarity criteria)

Statement: If the corresponding sides of two triangles are proportional then their corresponding angles are equal ,and hence the two triangles are similar

Given: Two triangles ABC and DEF such that \frac{AB}{DE} =\frac{BC}{EF} =\frac{AC}{DF}

To prove: ΔABC∼ΔDEF

Construction: Let p and Q be points on DE and DF respectively such that DP=AB and DQ=AC .Join PQ Proof:  We have

\frac{AB}{DE} =\frac{AC}{DF}

∴          \frac{DP}{DE} =\frac{DQ}{DF}            {AB=DP and AC=DQ by construction}

⇒       PQ is parallel to EF                                       {By converse of thale’s theorem}

Therefore            ∠DPQ=∠E    and ∠DQP=∠F                        {Corresponding angles}

Thus , In Triangle DPQ and DEF , we have

∠DPQ=∠E    and ∠DQP=∠F

Therefore ,by AA-similarity criteria ,we have

ΔDPQ∼ΔDEF

\frac{DP}{DE} =\frac{PQ}{EF}

\frac{AB}{DE} =\frac{PQ}{EF}                      {DP=AB}

But,        \frac{AB}{DE} =\frac{BC}{EF}               {All sides of similar triangle are similar}

∴             \frac{PQ}{DE} =\frac{BC}{EF}

Thus , in triangles ABC and DPQ , we have

AB=DP     ,AC=DQ  and BC=PQ

Therefore ,by SSS criteria of congruency ,we have

ΔABC≅ΔDPQ

From (i) and (ii) ,we have

ΔABC≅ΔDPQ    and ΔDPQ∼ΔDEF

⇒                ΔABC∼ΔDPQ  and ΔDPQ∼ΔDEF {Every congruent triangles are similar}

Since            ΔABC∼ΔDPQ  and ΔDPQ∼ΔDEF

so,                  ΔABC∼ΔDEF

Lesson from above two theorems

• Two triangles are similar if their corresponding angles are equal i.e they are equiangular.
• Two triangles are similar if their corresponding sides are proportional

Theorem 6.5: ( SAS similarity criterion)

Statement: 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 .

Given: Two triangles ABC and DEF such that ∠A=∠D and \frac{AB}{DE} =\frac{AC}{DF}

To prove: ΔABC∼ΔDEF

Construction: Mark points p and Q on DE and DF respectively such that DP=AB and DQ=AC.Join PQ Proof: In triangles ABC and DPQ, we have

AB=DP,∠A=∠D, and AC=DQ

Therefore ,by SAS criterion of congruence ,we have

ΔABC≅ΔDPQ                                       ……(i)

\frac{AB}{DE} =\frac{AC}{DF}

\frac{DP}{DE} =\frac{DQ}{DF}               [By the converse of the thale’s theorem]

∴      ∠DPQ=∠E  and ∠DQP=∠F

Therefore ,by AAA-criteria of similarity ,we have

ΔDPQ∼ΔDEF                                                  ……….(ii)

From eq(i) and (ii) ,we get

ΔABC≅ΔDPQ and ΔDPQ∼ΔDEF

ΔABC∼ΔDPQ and ΔDPQ∼ΔDEF

Therefore ΔABC∼ΔDEF

Important point:If two triangles ABC and DEF are similar ,then ration of their corresponding sides are proportional to the corresponding perimeres(Perimeter).

If Δ ABC and ΔDEF are similar ,then

\frac{AB}{DE} =\frac{BC}{EF} =\frac{AC}{DF}      [Similarity]

\frac{AB}{DE} =\frac{BC}{EF} =\frac{AC}{DF}=\frac{AB+BC+CA}{DE+EF+FD}

I hope this post was helpful for you .Don’t forget to share your feedback in the comment section .

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