Triangles Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution
Exercise 6.1
  1. Fill in the blanks using the correct word given in brackets: (i) All circles are…
  2. Give two different examples of pair of similar figures and non similar figures…
  3. State whether the following quadrilaterals are similar or not:
Exercise 6.2
  1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)…
  2. E and F are points on the sides PQ and PR respectively of a PQR. For each of the…
  3. In Fig. 6.18, if LM || CB and LN || CD, prove that
  4. In Fig. 6.19, DE || AC and DF || AE. Prove that bf/fe = be/ec
  5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
  6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB…
  7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of…
  8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides…
  9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at…
  10. The diagonals of a quadrilateral ABCD intersect each other at the point O such…
Exercise 6.3
  1. State which pairs of triangles in Fig. are similar.Write the similarity…
  2. In Fig. 6.35, ODC ~ OBA, BOC = 125 and CDO = 70. Find DOC, DCO and OAB…
  3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at…
  4. In Fig. 6.36, qr/qs = qt/pr and 1 = 2. Show that PQS ~ TQR.
  5. S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~…
  6. In Fig. 6.37, if ABE ACD, show that ADE ~ ABC.
  7. In Fig. 6.38, altitudes AD and CE of ABCintersect each other at the point P.…
  8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects…
  9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M…
  10. CD and GH are respectively the bisectors of ACB and EGF in such a way that D…
  11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC…
  12. Sides AB and BC and median AD of a triangle ABC are respectively proportional…
  13. D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that…
  14. Sides AB and AC and median AD of a triangle ABC are respectively proportional…
  15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the…
  16. If AD and PM are medians of triangles ABC and PQR, respectively where ABC ~…
Exercise 6.4
  1. Let ABC ~ DEF and their areas be, respectively, 64 cm^2 and 121 cm^2 . If EF =…
  2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.…
  3. In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD…
  4. If the areas of two similar triangles are equal, prove that they are congruent…
  5. D, E and F are respectively the mid-points of sides AB, BC and CA of ABC. Find…
  6. Prove that the ratio of the areas of two similar triangles is equal to the…
  7. Prove that the area of an equilateral triangle described on one side of a square…
  8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC.…
  9. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles…
Exercise 6.5
  1. Sides of triangles are given below. Determine which of them are right triangles.…
  2. PQR is a triangle right angled at P and M is a point on QR such that PM QR. Show…
  3. In Fig. 6.53, ABD is a triangle right angled at Aand AC BD. Show that: (i) AB^2…
  4. ABC is an isosceles triangle right angled at C. Prove that ab^2 = 2ac^2…
  5. ABC is an isosceles triangle with AC = BC. If ab^2 = 2ac^2 prove that ABC is a…
  6. ABC is an equilateral triangle of side 2a. Find each of its altitudes…
  7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum…
  8. In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and…
  9. A ladder 10 m long reaches a window 8 m above theground. Find the distance of…
  10. A wire attached to a vertical pole of height 18 m is 24 m long and has a stack…
  11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per…
  12. Two poles of heights 6 m and 11 m stand on aplane ground. If the distance…
  13. D and E are points on the sides CA and CB respectively of a triangle ABC right…
  14. The perpendicular from A on side BC of aABC intersects BC at D such that DB = 3…
  15. In an equilateral triangle ABC, D is a point on side BC such that bd = 1/3 bc…
  16. In an equilateral triangle, prove that three times the square of one side is…
  17. Tick the correct answer and justify: InABC, AB = 63 cm, AC = 12 cm and BC = 6…
Exercise 6.6
  1. In Fig. 6.56, PS is the bisector of QPR of PQR. Prove that qs/sr = pq/pr…
  2. In Fig. 6.57, D is a point on hypotenuse AC of ABC, DM BC and DN AB. Prove that:…
  3. In Fig. 6.58, ABC is a triangle in which ABC 90 and AD CB produced. Prove that…
  4. In Fig. 6.59, ABC is a triangle in which ABC 90 and AD BC. Prove that ac^2 =…
  5. In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that: (i) ac^2 =…
  6. Prove that the sum of the squares of the diagonals of parallelogram is equal to…
  7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove…
  8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point…
  9. In Fig. 6.63, D is a point on side BC of ABC such that bd/cd = ab/ac Prove that…
  10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above…

Exercise 6.1
Question 1.

Fill in the blanks using the correct word given in brackets:

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

(ii) All squares are ___________. (similar, congruent)

(iii) All ____________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _________ and (b) their corresponding sides are ___________. (equal, proportional)


Answer:

The solutions of the fill ups are:

(i) Similar


(ii) Similar


(iii) Equilateral


(iv) (a) Equal


(b) Proportional



Question 2.

Give two different examples of pair of

(i) Similar figures

(ii) Non-similar figures


Answer:

Two examples of similar figures are:

(i) Two equilateral triangles with sides 1 cm and 2 cm



(ii) Two squares with sides 1 cm and 2 cm



Now two examples of non-similar figures are:


(i) Trapezium and square



(ii) Triangle and parallelogram




Question 3.

State whether the following quadrilaterals are similar or not:



Answer:

The given quadrilateral PQRS and ABCD are not similar because though their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.




Exercise 6.2
Question 1.

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)



Answer:

(i) Let us take EC = x cm

Given: DE || BC

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Now, using basic proportionality theorem, we get:


 = 



x = 


x = 2 cm


Hence, EC = 2 cm


(ii) Let us take AD = x cm


Given: DE || BC

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Now, using basic proportionality theorem, we get




x = 


Hence, AD = 2.4 cm


Question 2.

E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm


Answer:

(i)

Given :
PE = 3.9 cm,

EQ = 3 cm,

PF = 3.6 cm,

FR = 2.4 cm


Now we know,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR






Therefore, EF is not parallel to QR


(ii)

We know that,
Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR



Hence,


Therefore, EF is parallel to QR


(iii)

In this we know that,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR




Hence,



EF is parallel to QR


Question 3.

In Fig. 6.18, if LM || CB and LN || CD, prove that




Answer:



Given: LM ll CB and LN ll CD
From the given figure:
In ΔALM and ΔABC

LM || CB
Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

Now, using basic proportionality theorem that the corresponding sides will have same proportional lengths, we get:


 (i)



And in ΔALN and ΔACD corresponding sides will be proportional


Therefore,


 (ii)


From (i) and (ii), we obtain



Hence, Proved.


Question 4.

In Fig. 6.19, DE || AC and DF || AE. Prove that





Answer:





Given: 

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides 
into two distinct points, then the line divides those sides in proportion.

In triangle ABC, DE is parallel to AC

Therefore,

By Basic proportionality theorem

 ......(1)

In triangle BAE, DF is parallel to AE


Therefore,

By Basic proportionality theorem

 ...... (2)

From (1) and (2), we get

Hence, Proved.
 
Question 5.

In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.


Answer:

To Prove: EF ll QR

Given: In triangle POQ, DE parallel to OQ

Proof:

In triangle POQ, DE parallel to OQ

Hence,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

 (Basic proportionality theorem) (i)

Now,



Hence,


 (Basic proportionality theorem) (ii)


From (i) and (ii), we get



Therefore,


EF is parallel to QR (Converse of basic proportionality theorem)

Hence, Proved.
 
Question 6.

In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.Show that BC || QR.


Answer:

To Prove: BC ll QR

Given that in triangle POQ, AB parallel to PQ

Hence,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

 (Basic proportionality theorem)


Now, 


Therefore,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Using Basic proportionality theorem, we get:



From above equations, we get



BC is parallel to QR (By the converse of Basic proportionality theorem)

Hence, Proved.
 
Question 7.

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX)


Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ is parallel to BC.


To Prove: PQ bisects AC
Given: PQ ll BC and PQ bisects AB
Proof: 
According to Theorem 6.1: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Now, using basic proportionality theorem, we get




[As AP = PB coz P is the mid-point of AB]

AQ = QC


Or, Q is the mid-point of AC
Hence proved.


Question 8.

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX)


Answer:


To Prove: PQ ll BC
Given: P and Q are midpoints of AB and AC
Proof: 
Let us take the given figure in which PQ is a line segment which joins the mid-points P and Q of line AB and AC respectively

i.e., AP = PB and AQ = QC


We observe that,



And,



Therefore,



Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Hence, using basic proportionality theorem we get:


PQ parallel to BC
Hence, Proved.


Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that 


Answer:

The figure is given below:


Given: ABCD is a trapezium
AB ll CD
Diagonals intersect at O
To Prove = 

Construction: Construct a line EF through point O, such that EF is parallel to CD.


Proof:

In ΔADC, EO is parallel to CD
According to basic propotionality theorem, if a side is drawn parallel to any side of the triangle then the corresponding sides formed are propotional

Now, using basic proportionality theorem in ΔABD and ΔADC, we obtain

In ΔABD, OE is parallel to AB

So, using basic proportionality theorem in ΔEOD and ΔABD, we get


From (i) and (ii), we get

Therefore by cross multiplying we get,


Hence, Proved


Question 10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such thatShow that ABCD is a trapezium


Answer:

The quadrilateral ABCD is shown below, BD and AC are the diagonals.

Construction: Draw a line OE parallel to AB

Given: In ΔABD, OE is parallel to AB

To prove : ABCD is a trapezium
According to basic proportionality theorem, if in a triangle another line is drawn parallel to any side of triangle, 
then the sides so obtain are proportional to each other.

Now, using basic proportionality theorem in ΔDOE and ΔABD, we obtain


 ...(i)


It is given that,


 ...(ii)


From (i) and (ii), we get


 ...(iii)
Now for ABCD to be a trapezium AB has to be parallel of CD

Now From the figure we can see that If eq(iii) exists then,

EO || DC (By the converse of basic proportionality theorem)

Now if,

⇒ AB || OE || DC

Then it is clear that

⇒ AB || CD

Thus the opposite sides are parallel and therefore it is a trapezium.

Hence,

ABCD is a trapezium



Exercise 6.3
Question 1.

State which pairs of triangles in Fig. are similar.Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:













Answer:

(i) From the figure:

∠A = ∠P = 60°


∠B = ∠Q = 80°


∠C = ∠R = 40°


Therefore, ΔABC ΔPQR [By AAA similarity]

Now corresponding sides of triangles will be propotional,



(ii)
From the triangle,



Hence the corresponding sides are propotional

Thus the corresponding angles will be equal

The triangles ABC and QRP are similar to each other by SSS similarity


(iii) The given triangles are not similar because the corresponding sides are not proportional


(iv) In triangle MNL and QPR, we have

∠M = ∠Q = 70o

but 

Therefore, MNL and QPR are not similar.

(v) In triangle ABC and DEF, we have


AB = 2.5, BC = 3


∠A = 80o


EF = 6


DF = 5


∠F = 80o



And, 


∠B ≠ ∠F


Hence, triangle ABC and DEF are not similar


(vi) In triangle DEF, we have


∠D + ∠E + ∠F = 180o (Sum of angles of triangle)


70o + 80o + ∠F = 180o


∠F = 30o


In PQR, we have


∠P + ∠Q + ∠R = 180o


∠P + 80o + 30o = 180o


∠P = 70o


In triangle DEF and PQR, we have


∠D = ∠P = 70o


∠F = ∠Q = 80o


∠F = ∠R = 30o


Hence, ΔDEF ~ ΔPQR (AAA similarity)


Question 2.

In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB


Answer:

From the figure,

We see, DOB is a straight line

∠ DOC + ∠ COB = 180° (angles on a straight line form a supplementary pair)

∠ DOC = 180° - 125°

∠ DOC = 55°

Now, In ΔDOC,

∠ DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180°)

∠ DCO + 70° + 55° = 180°

∠ DCO = 55°

It is given that ΔODC ΔOBA

∠ OAB = ∠ OCD (Corresponding angles are equal in similar triangles)


Thus,  OAB = 55°


Question 3.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Show that 


Answer:


In ΔDOC and ΔBOA,

∠CDO = ∠ABO (Alternate interior angles as AB || CD)


∠DCO = ∠BAO (Alternate interior angles as AB || CD)


∠DOC = ∠BOA (Vertically opposite angles)


Therefore,

ΔDOC ∼ ΔBOA [ BY AAA similarity]

Now in similar triangles, the ratio of corresponding sides are proportional to each other. Therefore,

 ......... (Corresponding sides are proportional)

or,


Hence, Proved.


Question 4.

In Fig. 6.36,  and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.



Answer:



To Prove: Δ PQS ∼ Δ TQR

Given: In ΔPQR,

∠PQR = ∠PRQ

Proof: 

As∠PQR = ∠PRQ

PQ = PR [sides opposite to equal angles are equal] (i)

Given,

 by (1)


In Δ PQS and Δ TQR, we get

∠Q = ∠Q


Therefore,

By SAS similarity Rule which states that Triangles are similar if two sides in one triangle are in the 
same proportion to the corresponding sides in the other, and the included angle are equal. 

Δ PQS ∼ Δ TQR 

Hence, Proved.


Question 5.

S and T are points on sides PR and QR of Δ PQR such that ∠P = ∠RTS. Show that Δ RPQ ~ Δ RTS


Answer:


In ΔRPQ and ΔRST,

∠ RTS = ∠ QPS (Given)

∠ R = ∠ R (Common to both the triangles)
If two angles of two triangles are equal, third angle will also be equal. As the sum of interior angles of triangle is constant and is 180°

∴ ΔRPQ ΔRTS (By AAA similarity)


Question 6.

In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.



Answer:

To Prove: Δ ADE ∼ Δ ABC

Given: ΔABE ≅ ΔACD

Proof: 

ΔABE ≅ ΔACD

∴ AB = AC (By CPCT) (i)

And,

AD = AE (By CPCT) (ii)



In ΔADE and ΔABC,


Dividing equation (ii) by (i)


∠A = ∠A (Common)

SAS Similarity: Triangles are similar if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angle are equal. 

Therefore,

ΔADEΔABC (By SAS similarity)

Hence, Proved.


Question 7.

In Fig. 6.38, altitudes AD and CE of Δ ABCintersect each other at the point P. Showthat:

(i) Δ AEP ~ Δ CDP

(ii) Δ ABD ~ Δ CBE

(iii) Δ AEP ~ Δ ADB

(iv) Δ PDC ~ Δ BEC



Answer:

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)


∠APE = ∠CPD (Vertically opposite angles)


Hence, by using AA similarity,


ΔAEP  ΔCDP


(ii) In ΔABD and ΔCBE,


∠ADB = ∠CEB (Each 90°)


∠ABD = ∠CBE (Common)


Hence, by using AA similarity,


ΔABD  ΔCBE


(iii) In ΔAEP and ΔADB,


∠AEP = ∠ADB (Each 90°)


∠PAE = ∠DAB (Common)


Hence, by using AA similarity,


ΔAEP  ΔADB


(iv) In ΔPDC and ΔBEC,


∠PDC = ∠BEC (Each 90°)


∠PCD = ∠BCE (Common angle)


Hence, by using AA similarity,


ΔPDC  ΔBEC



Question 8.

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB


Answer:



To Prove: Δ ABE ∼ Δ CFB
Given: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. As shown in the figure.
Proof:
In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram are equal)

∠AEB = ∠CBF (Alternate interior angles are equal because AE || BC)

Therefore,

ΔABE  ΔCFB (By AA similarity)

Hence, Proved.
 
Question 9.

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP

(ii) 



Answer:

(i) 

To Prove: Δ ABC ∼ Δ AMP

Given: In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

Proof:

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ΔABC  ΔAMP (By AA similarity)

Hence, Proved.


(ii) Δ ABC ∼ Δ AMP

Now we get that,

Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar. And the converse is also true, so we have


Hence, Proved.
 
Question 10.

CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF in such a way that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:
(i) 

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF


Answer:


Given, Δ ABC ∼ Δ FEG …..eq(1)


⇒ corresponding angles of similar triangles


⇒ ∠ BAC = ∠ EFG ….eq(2)


And ∠ ABC = ∠ FEG …….eq(3)


⇒ ∠ ACB = ∠ FGE


⇒ 


⇒ ∠ ACD = ∠ FGH and ∠ BCD = ∠ EGH ……eq(4)


Consider Δ ACD and Δ FGH


⇒ From eq(2) we have


⇒ ∠ DAC = ∠ HFG


⇒ From eq(4) we have


⇒ ∠ ACD = ∠ EGH


Also, ∠ ADC = ∠ FGH


⇒ If the 2 angle of triangle are equal to the 2 angle of another triangle, then by angle sum property of triangle 3rd angle will also be equal.


⇒ by AAA similarity we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar.


∴ Δ ADC ∼ Δ FHG


⇒ By Converse proportionality theorem


⇒ 


Consider Δ DCB and Δ HGE


From eq(3) we have


⇒ ∠ DBC = ∠ HEG


⇒ From eq(4) we have


⇒ ∠ BCD = ∠ FGH


Also, ∠ BDC = ∠ EHG


∴ Δ DCB ∼ ΔHGE


Hence proved.


Question 11.

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC,prove that Δ ABD ~ Δ ECF


Answer:



To Prove: Δ ABD ∼ Δ ECF


Given: ABC is an isosceles triangle, AD is perpendicular to BC

BC is produced to E and EF is perpendicular to AC

Proof: 


Given that ABC is an isosceles triangle

AB = AC

⇒ ∠ABD = ∠ECF


In ΔABD and ΔECF,


∠ADB = ∠EFC (Each 90°)


∠ABD = ∠ECF (Proved above)

Therefore,

ΔABD  ΔECF (By using AA similarity criterion)

AA Criterion: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

Hence, Proved.


Question 12.

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of
Δ PQR (see Fig. 6.41). Show that
Δ ABC ~ Δ PQR.


Answer:


To Prove: Δ ABC ∼ Δ PQR

Given:


Proof:
Median divides the opposite side

BD =  and,


QM = 


Now, 

Multiplying and dividing by 2, we get,

 = 



In Δ ABD and Δ PQM,


Side-Side-Side (SSS) Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar.

ΔABD  ΔPQM (By SSS similarity)


∠ABD = ∠PQM (Corresponding angles of similar triangles)


In ΔABC and ΔPQR,


∠ABD = ∠PQM (Proved above)


The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.

ΔABC  ΔPQR (By SAS similarity)

Hence, Proved.
 
Question 13.

D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB.CD 


Answer:

In ΔADC and ΔBAC,



To Prove : CA2 = CB . CD

Given: ∠ADC = ∠BAC 

Proof: Now In ΔADC and ΔBAC,

∠ADC = ∠BAC 

∠ACD = ∠BCA (Common angle)

According to AA similarity, if two corresponding angles of two triangles are equal then the triangles are similar

ΔADC  ΔBAC (By AA similarity)


We know that corresponding sides of similar triangles are in proportion


Hence in ΔADC and ΔBAC,





Hence Proved


Question 14.

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR.
Show that Δ ABC ~ Δ PQR


Answer:

To Prove: Δ ABC ∼ Δ PQR
Given:

Proof: 


Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. 
Then, join B to E, C to

E, Q to L, and R to L

We know that medians divide opposite sides.

Hence, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC,

Diagonals AE and BC bisect each other at point D.

Therefore,

Quadrilateral ABEC is a parallelogram.

AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given in the question that,