# Triangle class 10 theorems

## Triangle for class 10

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 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.

Example.

(i) Any two squares are always similar(because they have the same shape)but need not be congruent(Because for congruent, shape and size of both triangles should be equal)

(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 are not familiar with polygon, don’t worry I explain.

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

Example.Triangle ,square ,pentagon,hexagon,rectangle ,rhombus etc.Let discuss its criteria for similarity.

Two polygons are said to be similar, if

(i) Their corresponding angle 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.

Here are some problem based on the above concept

(i) All circles are………….(congruent, similar).

(ii)All…….triangles are similar(isosceles, equilaterals)

(iii) Two triangles are similar if their corresponding sides are ……..(proportional, equal)

True/False

(i) Any two similar figures are congruent.

(ii) Any two congruent figures are similar

(i) Similar

(ii) equilateral triangles

(iii) proportional

True/false

(i) True

(ii) False

### Triangle class 10 theorems

Here are some triangle class 10 theorems which you must practice

Theorem1 (Basic proportionality theorem or Thales theorem)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 intersect 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 same base DE and between 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 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 2:(converse of basic proportionality thorem 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 parallel to BC.Let DF\parallel BC

Since DF\parallel BC.So, by thales theorem

But it is given

Again LHS of both equation are equal

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

Here are some major which you should always keep in mind while proving

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

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

\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

### 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 side in the ratio of the sides containing the angle.e

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 }

and                        ∠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

### Theorem 4 (Converse of angle bisector theorem)

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)

and,            ∠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.

Arithmetic progression for class 10

Polynomial for class 10

Linear equation for class 10

### Theorem 5(Basic proportionality theorem)

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 2(Converse of mid-point theorem)

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 thale’s theorem ,we obtain DE\parallel BC

## Triangle class10 similarity theorems

Here are 3 triangle class 10 similarity theorems which is very very important.

Theorem.1(AAA similarity criteria)

Theorem 2(SSS similarity criteria) If the corresponding sides of two triangles are proportional .Then they 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 triangles 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 are similar}

Since            ΔABC∼ΔDPQ  and ΔDPQ∼ΔDEF

so,                  ΔABC∼ΔDEF

From above two theorems ,we learn

• 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 3( SAS similarity criterion).

If two triangles, one pair of corresponding sides are proportional and the included angles are equal, 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

Threfore ,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

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

⇒     \frac{AB}{DE} =\frac{BC}{EF} =\frac{AC}{DF}=

More triangle class 10 theorems on its properties

Theorem.4 If two triangles are equiangilar ,prove that the ratio of the corresponding sides is same as the ratio of the corresponding medians.

Given : Two triangles ABC and DEF in which ∠A=∠D ,∠B=∠E ,∠C=∠F,AP and DQ are their medians

To prove: \frac{BC}{EF} =\frac{AP}{DQ}

Proof: It is given ,Both triangles are equiangular .SO it will also similar

∴                  ΔABC∼ΔDEF           (AAA similarity criteria)

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

\frac{AB}{DE} =\frac{2BP}{2EQ}  { BC=2BP and EF=2EQ (AP and DQ are median)}

\frac{AB}{DE} =\frac{BP}{EQ}                    …………..(ii)

NOw ,in ΔABP and ΔDFQ,we have

\frac{AB}{DE} =\frac{BP}{EQ}

and ∠B=∠E

So, by SAS-criteria of similarity ,we have

ΔABP∼ΔDEQ

\frac{AB}{DE} =\frac{AP}{DQ}                   …………..(iii)

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

\frac{BC}{EF} =\frac{AP}{DQ}

similarly we can prove

\frac{AC}{DF} =\frac{AP}{DQ}

So, \frac{AB}{DE} =\frac{BC}{EF} =\frac{AC}{DF} =\frac{AP}{DQ}

Theorem.5 If two triangles are equiangular ,prove that the ratio of the corresponding sides is same as the ratio of the corresponding angle bisector segments.

Given.Two triangles ABC and DEF in which ∠A=∠D ,∠B=∠E ,∠C=∠F and AX,DY are the bisectors of ∠A and ∠D respectively.SO ∠BAX=∠EDY and ∠XAC=∠YDF

To prove:\frac{BC}{EF} =\frac{AX}{DY}

Proof: Since equiangular triangles are similar

∴          ΔABC∼ΔDEF           (AAA similarity criteria)

\frac{AB}{DE} =\frac{BC}{EF}                                               …….(i)

In ΔABX and ΔDEY ,we have

∠B=∠E           (Given)

and,∠BAX=∠EDY     {Since ∠A=∠D ⇒\frac{1}{2} \angle A=\frac{1}{2} \angle D=,∠BAX=∠E}

SO, By AA-criteria of similarity ,we have

ΔABX∼ΔDEY

\frac{AB}{DE} =\frac{AX}{DY}                                            ……..(ii)

from (i) and (ii) ,we get

\frac{BC}{EF} =\frac{AX}{DY}

Theorem.6 If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio,prove that the triangles are similar

Given: Two triangles ABC and DEF in which ∠A=∠D.The bisectors AP and DQ or ∠A and ∠D intersects BC and EF in P and Q respectively such that \frac{BP}{PC} =\frac{EQ}{QF}

To prove:      ΔABC∼ΔDEF

Proof: We know that the bisectors of an angle of a triangle intersects the oposite side in the same ratio

Since AP is the bisectors of ∠A

⇒      \frac{BP}{PC} =\frac{AB}{AC}       …………..(i)

Again,DQ is the bisector of ∠D

⇒      \frac{EQ}{QF} =\frac{DE}{DF}          ………….(ii)

But ,  \frac{BP}{PC} =\frac{EQ}{QF}

Therefore ,from (i) and (ii) ,we get

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

Thus,in ΔABC and ΔDEF ,we have

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

and, ∠A=∠D

So,by SAS- criteria of similarity ,we obtain

ΔABC∼ΔDEF