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). triangle

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

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. trapezium

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.
  trapezium
  Proof : We have AB || DC and EF || AB   (Given)
          so, EF || DC                    (Lines parallel to the same line are parallel to each other) 
          
          Now, in Δ ADC, 
          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. 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. triangle

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. triangle

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. triangle

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. triangle

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 triangle

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. triangle

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. triangle

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)
          So, AC/AB = AD/AC
          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),
          BC2/AC2 = (BA . BD)/(AB . AD) = BD/AD
    

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 
  triangle
  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. triangle

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. triangle

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. rectangle

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

          By Adding (i) and (ii),
          OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2 
          OB2 + OD2 = CQ2 + OP2 + OQ2 + AP2        (As BP = CQ and DQ = AP) 
          OB2 + OD2 = CQ2 + OQ2 + OP2 + AP2 
          OB2 + OD2 = OC2 + OA2 [From (iii) and (iv)]
    
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