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TOPICS
Unit-6(Examples)

Example-1 :-  If a line intersects sides AB and AC of a Δ ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC (see figure).

Solution :-
```Given that : DE || BC.
Prove that : AD/AB = AE/EC
Proof : We have DE || BC                (Given)
Both sides adding by 1, we get
DB/AD + 1 = EC/AE + 1
```

Example-2 :-  ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Figure). Show that AE/ED = BF/FC.

Solution :-
```  Given that : ABCD is a trapezium with AB || DC and EF || AB.
Prove that : AE/ED = BF/FC.
Construction : Let us join AC to intersect EF at G.

Proof : We have AB || DC and EF || AB   (Given)
so, EF || DC                    (Lines parallel to the same line are parallel to each other)

EG || DC                        (As EF || DC)
So, AE/ED = AG/GC ......(i)     (Theorem-1)
Similarly, from Δ CAB,
CG/AG = CF/BF or
AG/GC = BF/FC   .......(ii)
Therefore, from (i) and (ii)
AE/ED = BF/FC
```

Example-3 :-  In figure, PS/SQ = PT/TR and ∠ PST = ∠ PRT. Prove that PQR is an isosceles triangle.

Solution :-
```  Given that : PS/SQ = PT/TR and ∠ PST = ∠ PRQ.
Prove that : PQR is an isosceles triangle.
Proof : We have PS/SQ = PT/TR           (Given)
so, ST || QR                    (Theorem-2)
Therefore, ∠ PST = ∠ PQR....(i) (Corresponding angles)
Also, we have
∠ PST = ∠ PRQ ......(ii)        (Given)
So, ∠ PRQ = ∠ PQR [From (i) and (ii)]
Therefore, PQ = PR              (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle.
```

Example-4 :-  In Figure, if PQ || RS, prove that Δ POQ ~ Δ SOR.

Solution :-
```  Given that : PQ || RS
Prove that : Δ POQ ~ Δ SOR
Proof : We have PQ || RS                (Given)
So, ∠ P = ∠ S                   (Alternate angles)
and ∠ Q = ∠ R
Also, ∠ POQ = ∠ SOR             (Vertically opposite angles)
Therefore, Δ POQ ~ Δ SOR        (AAA similarity criterion)
```

Example-5 :-  Observe Figure and then find ∠ P.

Solution :-
```   In Δ ABC and Δ PQR,
AB/RQ = 3.8/7.6 = 1/2
BC/QP = 6/12 = 1/2
CA/PR = 3√3/6√3 = 1/2
i.e. AB/RQ = BC/QP = CA/PR
So, Δ ABC ~ Δ RQP                       (SSS similarity)
Therefore, ∠ C = ∠ P                    (Corresponding angles of similar triangles)
But ∠ C = 180o – ∠ A – ∠ B              (Angle sum property)
= 180o – 80o – 60o
= 40o
So, ∠ P = 40o
```

Example-6 :-  In Figure, OA . OB = OC . OD. Show that ∠ A = ∠ C and ∠ B = ∠ D.

Solution :-
```  Given that : OA . OB = OC . OD
Prove that : ∠ A = ∠ C and ∠ B = ∠ D
Proof : We have  OA . OB = OC . OD      (Given)
So, OA/OC = OD/OB ......(i)     (Alternate angles)
Also, we have ∠ AOD = ∠ COB .....(ii)   (Vertically opposite angles)
Therefore, from (i) and (ii),
Δ AOD ~ Δ COB                   (SAS similarity criterion)
So, ∠ A = ∠ C and ∠ D = ∠ B     (Corresponding angles of similar triangles)
```

Example-7 :-  A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Solution :-
```  Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post.
In figure, DE is the shadow of the girl. Let DE be x metres.
Now, BD = 1.2 m × 4 = 4.8 m.
In Δ ABE and Δ CDE,
∠ B = ∠ D           (Each is of 90o because lamp-post as well as the girl are standing vertical to the ground)
and ∠ E = ∠ E       (Same angle)
So, Δ ABE ~ Δ CDE   (AA similarity criterion)
Therefore, (4.8 + x)/x = 3.6/0.9
4.8 + x = 4x
4.8 = 4x - x
4.8 = 3x
4.8/3 = x
x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
```

Example-8 :-  In Figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that :
(i) Δ AMC ~ Δ PNR
(ii) CM/RN = AB/PQ
(iii) Δ CMB ~ Δ RNQ

Solution :-
```Given that : CM and RN are respectively the medians of Δ ABC and Δ PQR. Δ ABC ~ Δ PQR
Prove that :(i) Δ AMC ~ Δ PNR
(ii) CM/RN = AB/PQ
(iii) Δ CMB ~ Δ RNQ
Proof : (i) We have  Δ ABC ~ Δ PRQ                  (Given)
So, AB/PQ = BC/QR = CA/RP ......(i)
and ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R .....(ii)
AB = 2 AM and PQ = 2 PN                 (As CM and RN are medians)
So, from (i), 2AM/2PN = CA/RP
i.e., AM/PN = CA/RP  .......(iii)
Also, ∠ MAC = ∠ NPR [From (ii)] ......(iv)
So, from (iii) and (iv),
Δ AMC ~ Δ PNR .......(v)                (SAS similarity)
```
```          (ii) From (v) CM/RN = CA/RP ....(vi)
CA/RP = AB/PQ   [from (i)]  ....(vii)
Therefore, CM/RN = AB/PQ  [from (vi) and (vii)] .....(viii)
```
```          (iii) Again, AB/PQ = BC/QR [frrom (i)]
Therefore, CM/RN = BC/QR  [from (viii)] .....(ix)
Also, CM/RN = AB/PQ = 2BM/2QN
CM/RN = BM/QN  .......(x)
CM/RN = BC/QR = BM/QN     [from (ix) and (x)]
Therefore, Δ CMB ~ Δ RNQ                (SSS similarity)
```

Example-9 :-  In Figure, the line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas. Find the ratio AX/AB.

Solution :-
```  Given that The line segment XY is parallel to side AC of Δ ABC.
i.e., XY || AC
So, ∠ BXY = ∠ A and ∠ BYX = ∠ C                  (Corresponding angles)
Therefore, Δ ABC ~ Δ XBY                         (By AA similarity Rule)
So, ar(ABC)/ar(XBY) = (AB/XB)2....(i) (Therorem-6)

Also given It divides the triangle into two parts of equal areas.
i.e., ar (ABC) = 2 ar (XBY)
So, ar(ABC)/ar(XBY) = 2/1  .........(ii)
Therefore, from (i) and (ii),
(AB/XB)2 = 2/1
AB/XB = √2/1
XB/AB = 1/√2
1 - XB/AB = 1 - 1/√2
(AB - XB)/AB = (√2 - 1)/√2
AX/AB = (2 - √2)/2
```

Example-10 :-  In Figure, ∠ ACB = 90o and CD ⊥ AB. Prove that BC2/AC2 = BD/AD.

Solution :-
```  Given that : ∠ ACB = 90o and CD ⊥ AB.
Prove that : BC2/AC2 = BD/AD
Proof : We have ∠ ACB = 90o and CD ⊥ AB         (Given)
Δ ACD ~ Δ ABC                           (Theorem-7)
AC2 = AB . AD ......(i)
Similarly, Δ BCD ~ Δ BAC                (Theorem-7)
So, BC/BA = BD/BC
BC2 = BA . BD ......(ii)
Therefore, from (i) and (ii),
```

Example-11 :-  A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Solution :-
```  Let AB be the ladder and CA be the wall with the window at A.
Also, BC = 2.5 m and CA = 6 m

From Pythagoras Theorem, we have:
AB2 = BC2 + CA2
AB2 = (2.5)2 + (6)2
AB2 = 42.25
AB = 6.5
Thus, length of the ladder is 6.5 m.
```

Example-12 :-  In Figure, if AD ⊥ BC, prove that AB2 + CD2 = BD2 + AC2.

Solution :-
```  Given that : AD ⊥ BC
Prove that : AB2 + CD2 = BD2 + AC2
Proof : In Δ ADC, we have AC2 = AD2 + CD2 ....(i) (By Pythagoras Theorem)
In Δ ADB, we have AB2 = AD2 + BD2 ....(ii)(By Pythagoras Theorem)

By Subtracting (i) from (ii), we get
AB2 – AC2 = BD2 – CD2
AB2 + CD2 = BD2 + AC2
```

Example-13 :-  BL and CM are medians of a triangle ABC right angled at A. Prove that 4(BL2 + CM2) = 5BC2.

Solution :-
```  Given that : BL and CM are medians of the Δ ABC in which ∠ A = 90o
Prove that : 4(BL2 + CM2) = 5BC2
Proof : In Δ ABC, we have BC2 = AB2 + AC2 ....(i) (By Pythagoras Theorem)
In Δ ABL, we have BL2 = AL2 + AB2         (By Pythagoras Theorem)
BL2 = (AC/2)2 + AB2      (L is the mid point of AC)
BL2 = AC2/4 + AB2
4BL2 = AC2 + 4AB2  .......(ii)

In Δ CMA, we have CM2 = AC2 + AM2         (By Pythagoras Theorem)
CM2 = AC2 + (AB/2)2     (M is the mid point of AB)
CM2 = AC2 + AB2/4
4CM2 = 4AC2 + AB2  .......(iii)

By Adding (ii) and (iii), we get
4(BL2 + CM2) = 5(AC2 + AB2)
4(BL2 + CM2) = 5BC2 [From (i)]
```

Example-14 :-  O is any point inside a rectangle ABCD (see Figure). Prove that OB2 + OD2 = OA2 + OC2.

Solution :-
```  Given that : O is any point inside a rectangle ABCD.
Prove that : OB2 + OD2 = OA2 + OC2
Construction : Draw PQ || BC

Proof : Through O, draw PQ || BC so that P lies on AB and Q lies on DC.
Now, PQ || BC
Therefore, PQ ⊥ AB and PQ ⊥ DC(∠ B = 90o and ∠ C = 90o)
So, ∠ BPQ = 90o and ∠ CQP = 90o
Therefore, BPQC and APQD are both rectangles.
Now, in Δ OPB, OB2 = BP2 + OP2 ...... (i)
Similarly, in Δ OQD, OD2 = OQ2 + DQ2 ...... (ii)
Similarly, in Δ OQC, OC2 = OQ2 + CQ2 ...... (iii)
Similarly, in Δ OAP, OA2 = AP2 + OP2 ...... (iv)