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TOPICS
Exercise - 6.3

Question-1 :-  State which pairs of triangles in Figure 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 : Solution :-
```(i) In Δ ABC & Δ PQR
∠A = ∠P = 60o
∠B = ∠Q = 80o
∠C = ∠R = 40o
Therefore, Δ ABC ~ Δ PQR (By AAA similarity rule)
Hence, AB/QR = BC/RP = CA/PQ
```
```(ii) In Δ ABC & Δ PQR
AB = 2, QR = 4
BC = 2.5, PR = 5
AC = 3, PQ = 6
By using proportional rule, we get
QR/AB = PR/BC = PQ/AC
4/2 = 5/2.5 = 6/3 = 2
Therefore, Δ ABC ~ Δ PQR (By SSS similarity rule)
```
```(iii) Both triangles are not similar as the corresponding sides are not proportional.
```
```(iv) In Δ LMN & Δ PQR
PQ = 5, QR = 10
MN = 2.5, ML = 5
∠M = ∠Q = 70o
By using Proportional rule, we get
PQ/MN = QR/ML
5/2.5 = 10/5 = 2
Therefore, Δ LMN ~ Δ PQR (By SAS similarity rule)
```
```(v) Both triangles are not similar as the corresponding sides are not proportional.
```
```(vi) In Δ DEF,
∠D +∠E +∠F = 180o           (Sum of the measures of the angles of a triangle is 180o.)
70o + 80o + ∠F = 180o
∠F = 30o

Similarly, in Δ PQR,
∠P +∠Q + ∠R = 180o           (Sum of the measures of the angles of a triangle is 180o.)
∠P + 80o + 30o = 180o
∠P = 70o

In Δ DEF and Δ PQR,
∠D = ∠P (Each 70o)
∠E = ∠Q (Each 80o)
∠F = ∠R (Each 30o)
Therefore, ΔDEF ∼ ΔPQR      (By AAA similarity criterion)

```

Question-2 :-  In Figure, Δ ODC ~ Δ OBA, ∠ BOC = 125o and ∠ CDO = 70o. Find ∠ DOC, ∠ DCO and ∠ OAB. Solution :-
```    Here, DOB is a straight line.
So, ∠ DOB = 180o,
∠ BOC = 125o                (Given)
Now, ∠ DOB = ∠ DOC + ∠ BOC
180o = ∠ DOC + 125o
∠ DOC = 180o - 125o
∠ DOC = 55o

In Δ DOC, Given that ∠ CDO = 70o
∠ DOC + ∠ DCO + ∠ CDO = 180o    (Sum of the measures of the angles of a triangle is 180o)
55o + ∠ DCO + 70o = 180o
∠ DCO = 180o - 125o
∠ DCO = 55o

In figure, Given that Δ ODC ~ Δ OBA
So, ∠OAB = ∠OCD                         (Corresponding angles are equal in similar triangles)
∠OAB = 55o
```

Question-3 :-  Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

Solution :-
``` Given that : Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O.
Prove that : OA/OC = OB/OD
Proof : In Δ DOC and Δ BOA,
∠CDO = ∠ABO             (Alternate interior angles as AB || CD)
∠DCO = ∠BAO             (Alternate interior angles as AB || CD)
∠DOC = ∠BOA             (Vertically opposite angles)

By AAA similarity criterion Rule
Therefore Δ DOC ∼ Δ BOA
Hence, OD/OB = OC/OA    (Corresponding sides are Proportional)
OA/OC = OB/OD
```

Question-4 :-  In Figure, QR/QS = QT/PR and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR. Solution :-
```Given that : QR/QS = QT/PR and ∠1 = ∠2.
Prove that : Δ PQS ~ Δ TQR
Proof : In Δ PQR,
∠1 = ∠2 or ∠PQR = ∠PRQ     (Given)
Therefore, PQ = PR .......(i)
Also, given that QR/QS = QT/PR
By using (i), we get
QR/QS = QT/QP ......(ii)

In Δ PQS and Δ TQR
QR/QS = QT/QP
By using (ii), we get
∠1 = ∠1
By SAS similarity rule, we get
Δ PQS ~ Δ TQR
```

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.

Solution :-
``` Given that : S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS.
Proove that : Δ RPQ ~ Δ RTS
Proof : In ΔRPQ and ΔRST,
∠RTS = ∠QPS                     (Given)
∠PRQ = ∠TRS                     (Common angle)
Therefore, By AA similarity rule
ΔRPQ ∼ ΔRTS
```

Question-6 :-  In Figure, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC. Solution :-
```Given that : Δ ABE ≅ Δ ACD
Prove that : Δ ADE ~ Δ ABC
Proof : Δ ABE ≅ Δ ACD                    (Given)
Therefore, AB = AC .....(i)      (By CPCT Rule)
And, AD = AE ......(ii)          (By CPCT Rule)
AD/AB = AE/EC                    (By Dividing equation (ii) by (i))
∠QBAC = ∠QDAE                   (Common angle)
Therefore, By SAS similarity Rule,
```

Question-7 :-  In Figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:
(i) Δ AEP ~ Δ CDP
(ii) Δ ABD ~ Δ CBE
(iii) Δ AEP ~ Δ ADB
(iv) Δ PDC ~ Δ BEC Solution :-
```Given that : Altitudes AD and CE of Δ ABC intersect each other at the point P.
Prove that : (i) Δ AEP ~ Δ CDP
(ii) Δ ABD ~ Δ CBE
(iii) Δ AEP ~ Δ ADB
(iv) Δ PDC ~ Δ BEC
Proof :

(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP                 (Each 90o)
∠APE = ∠CPD                 (Vertically opposite angles)
Therefore, By using AA similarity rule,
ΔAEP ∼ ΔCDP
```
```Proof :

(ii) In ΔABD and ΔCBE,
∠ABD = ∠CBE                 (Common angle)
Therefore, By using AA similarity rule,
ΔABD ∼ ΔCBE
```
```Proof:

∠PAE = ∠DAB                 (Common angle)
Therefore, By using AA similarity rule,
```
```Proof :

(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC                 (Each 90o)
∠PCD = ∠BCE                 (Common angle)
Therefore, By using AA similarity rule,
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.

Solution :-
``` Given that : E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
Prove that : Δ ABE ~ Δ CFB
Proof : In ΔABE and ΔCFB,
∠BAE = ∠BCF                 (Opposite angles of a parallelogram)
∠AEB = ∠CBF                 (Alternate interior angles as AE || BC)
Therefore, By AA similarity rule,
ΔABE ∼ ΔCFB
```

Question-9 :-  In Figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP
(ii) CA/PA = BC/MP Solution :-
```Given that : ABC and AMP are two right triangles, right angled at B and M respectively.
Prove that : (i) Δ ABC ~ Δ AMP
(ii) CA/PA = BC/MP
Proof :

(i) In ΔABC and ΔAMP,
∠ABC = ∠AMP                 (Each 90o)
∠BAC = ∠MAP                 (Common angle)
Therefore, By AA similarity rule
ΔABC ∼ ΔAMP
```
```Proof :

(ii) ΔABC ∼ ΔAMP, So
CA/PA = BC/MP               (Corresponding sides of similar triangles are proportional)
```

Question-10 :-  CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such 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) CD/GH = AC/FG
(ii) Δ DCB ~ Δ HGE
(iii) Δ DCA ~ Δ HGF

Solution :-
``` Given that : CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF
such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively.
Δ ABC ~ Δ FEG
Prove that : (i) CD/GH = AC/FG
(ii) Δ DCB ~ Δ HGE
(iii) Δ DCA ~ Δ HGF
Proof :

(i) Δ ABC ~ Δ FEG        (Given)
Therefore, ∠BAC = ∠EFG, ∠ABC = ∠FEG and ∠ACB = ∠FGE
∠ACD = ∠FGH         (Angle Bisectors)
∠DCB = ∠HGE         (Angle Bisectors)

In Δ ACD and Δ FGH,
∠BAC = ∠FEG         (Δ ABC ~ Δ FEG)
∠ACD = ∠FGH         (Above Proved)
CD/GH = AC/FG
```
``` Proof :

(ii) In Δ DCB and Δ HGE,
∠DCB = ∠HGE         (Proved above)
∠ABC = ∠FEG         (Proved above)
Δ DCB ∼ Δ HGE       (By AA similarity rule)
```
``` Proof :

(iii) In Δ DCA and Δ HGF,
∠ACD = ∠FGH         (Proved above)
∠BAC = ∠EFG         (Proved above)
Δ DCA ∼ Δ HGF       (By AA similarity rule)
```

Question-11 :-  In Figue, 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. Solution :-
```Given that : E is a point on side CB produced of an isosceles triangle ABC with AB = AC. Also, AD ⊥ BC and EF ⊥ AC
Prove that : Δ ABD ~ Δ ECF
Proof : Δ ABC is an isosceles triangle.     (Given)
Therefore, AB = AC, ∠ABD = ∠ECF
In Δ ABD and Δ ECF,
Δ ABD ∼ Δ ECF        (By using AA similarity criterion)
```

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 Figure). Show that Δ ABC ~ Δ PQR. Solution :-
``` Given that : 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 .
Prove that : Δ ABC ~ Δ PQR
Proof : Sides AB and BC and median AD of a Δ ABC.
Median divides the opposite side.
Therefore, BD = BC/2 and QM = QR/2

Also given that

In Δ ABD and Δ PQM,
AB/PQ = BD/QM = AD/PM   (Proved above)
Δ ABD ∼ ΔPQM            (By SSS similarity rule)
∠ABD = ∠PQM             (Corresponding angles of similar triangles)

In Δ ABC and Δ PQR,
∠ABD = ∠PQM             (Proved above)
AB/PQ = BC/QR
Δ ABC ∼ Δ PQR           (By SAS similarity rule)
```

Question-13 :-  D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB . CD.

Solution :-
``` Given that : D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC.
Prove that : CA2 = CB . CD
Proof : In Δ ADC and Δ BAC,
∠ACD = ∠BCA             (Common angle)
Δ ADC ∼ Δ BAC           (By AA similarity rule)
We know that corresponding sides of similar triangles are in proportion.
Therefore, CA/CB = CD/CA
CA2 = CB . CD
```

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.

Solution :-
``` Given that : AB/PQ = AC/PR = AD/PM
Prove that : Δ ABC ~ Δ PQR
Construction : 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. Proof : We know that medians divide opposite sides.
Therefore, 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.
So, 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 that AB/PQ = AC/PR = AD/PM
AB/PQ = BE/QL = AE/PL
Therefore, Δ ABE ∼ Δ PQL   (By SSS similarity rule)

We know that corresponding angles of similar triangles are equal.
Therefore, ∠BAE = ∠QPL ....... (i)
Similarly, it can be proved that Δ AEC ∼ Δ PLR and ∠CAE = ∠RPL ...... (ii)
Adding equation (i) and (ii), we get
∠BAE + ∠CAE = ∠QPL + ∠RPL
∠CAB = ∠RPQ .......... (iii)

In Δ ABC and Δ PQR,
AB/PQ = AC/PR               (Given)
∠CAB = ∠RPQ                 (Using equation (iii))
Δ ABC ∼ Δ PQR               (By SAS similarity rule)
```

Question-15 :-  A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution :-
``` Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, ∠DCF = ∠BAE
And, ∠DFC = ∠BEA
∠CDF = ∠ABE                     (Tower and pole are vertical to the ground)
ΔABE ∼ ΔCDF                     (AAA similarity criterion)
AB/CD = BE/DF
AB/6 = 28/4
AB = 42 m
Therefore, the height of the tower will be 42 metres.
```

Question-16 :-  If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that AB/PQ = AD/PM.

Solution :-
``` Given that : AD and PM are medians of triangles ABC and PQR and Δ ABC ∼ Δ PQR.
Prove that : AB/PQ = AD/PM
Proof : Δ ABC ∼ Δ PQR               (Given)
We know that the corresponding sides of similar triangles are in proportion.
Therefore, AB/PQ = AC/PR = BC/QR ....... (i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ....... (ii)
Since AD and PM are medians, they will divide their opposite sides.
Therefore, BD = BC/2, QM = QR/2 .......(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ......... (iv)

In Δ ABD and Δ PQM,
∠B = ∠Q                     (Using equation (ii))
AB/PQ = BD/QM               (Using equation(iv))
Δ ABD ∼ Δ PQM               (By SAS similarity rule)