TOPICS
Exercise - 6.6

Question-1 :-  In Figure, PS is the bisector of ∠ QPR of Δ PQR. Prove that QS/SR = PQ/PR. triangle

Solution :-
Given that : In Δ PQR PS is the bisector of ∠ QPR of Δ PQR.
Prove that : QS/SR = PQ/PR
Construction : Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
    triangle
Proof : Given that, PS is the angle bisector of ∠QPR.
        ∠QPS = ∠SPR … (1) By construction,
        ∠SPR = ∠PRT (As PS || TR) ........ (i) 
        ∠QPS = ∠QTR (As PS || TR) ........ (ii)
        Using these equations, we obtain
        ∠PRT = ∠QTR 
        Therefore, PT = PR
        By construction, PS || TR
        By using basic proportionality theorem for ΔQTR,
        QS/SR = PQ/PT
        QS/SR = PQ/PR               (PT = TR)
    

Question-2 :-  In Figure, D is a point on hypotenuse AC of Δ ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that :
(i) DM2 = DN . MC
(ii) DN2 = DM . AN triangle

Solution :-
Given that : In Δ ABC, D is a point on hypotenuse AC and BD ⊥ AC, DM ⊥ BC and DN ⊥ AB.
Prove that : (i) DM2 = DN . MC 
            (ii) DN2 = DM . AN
Construction : Let us join DB.
    triangle
Proof : (i) We have, DN || CB, DM || AB, and ∠B = 90o
        Therefore, DMBN is a rectangle.
        Hence, DN = MB and DM = NB
        The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
        Therefore, ∠CDB = 90o 
        ∠2 + ∠3 = 90o ........ (i) 
        In Δ CDM,
        ∠1 + ∠2 + ∠DMC = 180o 
        ∠1 + ∠2 = 90o ........ (ii)
        In Δ DMB,
        ∠3 + ∠DMB + ∠4 = 180o  
        ∠3 + ∠4 = 90o ........ (iii)
        From equation (i) and (ii), we obtain ∠1 = ∠3
        From equation (i) and (iii), we obtain ∠2 = ∠4
        In Δ DCM and Δ BDM,
        ∠1 = ∠3             (Proved above) 
        ∠2 = ∠4             (Proved above)
        Δ DCM ∼ Δ BDM       (By AA similarity criterion)
        BM/DM = DM/MC
        DN/DM = DM/MC       (BM = DN)
        DM2 = DN . MC
    
Proof : (ii) ) In right Δ DBN, 
        ∠5 + ∠7 = 90o ........ (iv)
        In right Δ DAN, 
        ∠6 + ∠8 = 90o ........ (v)
        D is the foot of the perpendicular drawn from B to AC. 
        Therefore, ∠ADB = 90o
        ∠5 + ∠6 = 90o ........ (vi)
        From equation (iv) and (vi), we obtain ∠6 = ∠7
        From equation (v) and (vi), we obtain ∠8 = ∠5
        In Δ DNA and Δ BND,
        ∠6 = ∠7             (Proved above) 
        ∠8 = ∠5             (Proved above)
        Δ DNA ∼ Δ BND       (By AA similarity criterion)
        AN/DN = DN/NB
        AN/DN = DN/DM       (NB = DM)
        DN2 = AN . DM 
    

Question-3 :-  In Figure, ABC is a triangle in which ∠ ABC > 90o and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC . BD. triangle

Solution :-
Given that : ABC is a triangle in which ∠ ABC > 90o and AD ⊥ CB produced.
Prove that : AC2 = AB2 + BC2 + 2 BC . BD
Proof : Applying Pythagoras theorem in Δ ADB, we obtain AB2 = AD2 + DB2 ....... (i)
        Applying Pythagoras theorem in Δ ACD, we obtain AC2 = AD2 + DC2
        AC2 = AD2 + (DB + BC)2
        AC2 = AD2 + DB2 + BC2 + 2DB . BC
        By using equation (i), we get
        AC2 = AB2 + BC2 + 2DB . BC
    

Question-4 :-  In Figure, ABC is a triangle in which ∠ ABC < 90o and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2 BC . BD. triangle

Solution :-
Given that : ABC is a triangle in which ∠ ABC < 90o and AD ⊥ BC.
Prove that : AC2 = AB2 + BC2 – 2 BC . BD
Proof : Applying Pythagoras theorem in Δ ADB, we obtain AD2 + DB2 = AB2
        AD2 = AB2 − DB2 ........ (i)
        Applying Pythagoras theorem in Δ ADC, we obtain 
        AD2 + DC2 = AC2
        By using eqaution (i), we get
        AB2 − BD2 + DC2 = AC2
        AB2 − BD2 + (BC − BD)2 = AC2
        AC2 = AB2 − BD2 + BC2 + BD2 − 2 BC . BD 
        AC2 = AB2 + BC2 − 2 BC . BD
    

Question-5 :-  In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC . DM + (BC/2)2
(ii) AB2 = AD2 - BC . DM + (BC/2)2
(iii) AC2 + AB2 = 2AD2 + BC2/2 triangle

Solution :-
Given that : AD is a median of a Δ ABC and AM ⊥ BC.
Prove that : (i) AC2 = AD2 + BC . DM + (BC/2)2
            (ii) AB2 = AD2 - BC . DM + (BC/2)2
           (iii) AC2 + AB2 = 2AD2 + BC2/2
Proof : (i) Applying Pythagoras theorem in Δ AMD, we obtain 
        AM2 + MD2 = AD2 .......... (i)
        Applying Pythagoras theorem in Δ AMC, we obtain 
        AM2 + MC2 = AC2
        AM2 + (MD + DC)2 = AC2
        (AM2 + MD2) + DC2 + 2 MD . DC = AC2
        By using equation (i), we get
        AD2 + DC2 + 2 MD . DC = AC2 
        Using the result, DC = BC/2 , we obtain
        AD2 + (BC/2)2 + 2 MD(BC/2) = AC2
        AD2 + (BC/2)2 + MD . BC = AC2
    
Proof : (ii) Applying Pythagoras theorem in Δ ABM, we obtain 
        AB2 = AM2 + MB2
        AB2 = (AD2 − DM2) + MB2
        AB2 = (AD2 − DM2) + (BD − MD)2
        AB2 = AD2 − DM2 + BD2 + MD2 − 2 BD . MD 
        AB2 = AD2 + BD2 − 2 BD . MD
        AB2 = AD2 + (BC/2)2 - 2 x BC/2 x MD
        AB2 = AD2 + (BC/2)2 - BC . MD
    
Proof : (iii) Applying Pythagoras theorem in Δ ABM, we obtain 
        AM2 + MB2 = AB2 ....... (i)
        Applying Pythagoras theorem in Δ AMC, we obtain 
        AM2 + MC2 = AC2 ....... (ii)
        Adding equations (i) and (ii), we obtain 
        2AM2 + MB2 + MC2 = AB2 + AC2
        2AM2 + (BD − DM)2 + (MD + DC)2 = AB2 + AC2
        2AM2 + BD2 + DM2 − 2 BD . DM + MD2 + DC2 + 2 MD . DC = AB2 + AC2 
        2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2
        2(AM2 + MD2) + (BC/2)2 + (BC/2)2 + 2MD(-BC/2 + BC/2) = AB2 + AC2
        2AD2 + BC2/2 = AB2 + AC2
    

Question-6 :-  Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution :-
    triangle
    Let ABCD be a parallelogram.
    Let us draw perpendicular DE on extended side AB, and AF on side DC.
    Applying Pythagoras theorem in Δ DEA, we obtain 
    DE2 + EA2 = DA2 ........ (i)
    Applying Pythagoras theorem in Δ DEB, we obtain 
    DE2 + EB2 = DB2
    DE2 + (EA + AB)2 = DB2
    (DE2 + EA2) + AB2 + 2 EA . AB = DB2 
    DA2 + AB2 + 2 EA . AB = DB2 ......... (ii)
    Applying Pythagoras theorem in Δ ADF, we obtain 
    AD2 = AF2 + FD2
    Applying Pythagoras theorem in Δ AFC, we obtain 
    AC2 = AF2 + FC2
    AC2 = AF2 + (DC − FD)2
    AC2 = AF2 + DC2 + FD2 − 2 DC . FD 
    AC2 = (AF2 + FD2) + DC2 − 2 DC . FD
    AC2 = AD2 + DC2 − 2 DC . FD ........ (iii)
    Since ABCD is a parallelogram,
    AB = CD ........ (iv)
    BC = AD ........ (v)
    In Δ DEA and Δ ADF,
    ∠DEA = ∠AFD                 (Both 90o) 
    ∠EAD = ∠ADF                 (EA || DF) 
    AD = AD                     (Common side)
    Δ EAD ≅ Δ FDA               (By using AAS congruency criterion) 
    EA = DF ......... (vi)
    Adding equations (i) and (iii), we obtain
    DA2 + AB2 + 2 EA . AB + AD2 + DC2 − 2 DC . FD = DB2 + AC2 
    DA2 + AB2 + AD2 + DC2 + 2 EA . AB − 2 DC . FD = DB2 + AC2 
    BC2 + AB2 + AD2 + DC2 + 2 EA . AB − 2 AB . EA = DB2 + AC2 
    By Using equations (iv) and (vi), we get
    AB2 + BC2 + CD2 + DA2 = AC2 + BD2
    

Question-7 :-  In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) Δ APC ~ Δ DPB
(ii) AP . PB = CP . DP triangle

Solution :-
Given that : Two chords AB and CD intersect each other at the point P.
Prove that : (i) Δ APC ~ Δ DPB 
            (ii) AP . PB = CP . DP
Construction : Let us join CB.
    triangle
Proof : (i) In Δ APC and Δ DPB,
        ∠APC = ∠DPB             (Vertically opposite angles)
        ∠CAP = ∠BDP             (Angles in the same segment for chord CB) 
        Δ APC ∼ Δ DPB           (By AA similarity criterion)
    
Proof : (ii) We have already proved that Δ APC ∼ Δ DPB
        We know that the corresponding sides of similar triangles are proportional.
        Therefore, AP/DP = PC/PB = CA/BD
        AP/DP = PC/PB
        AP . PB = PC . DP
    

Question-8 :- In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) Δ PAC ~ Δ PDB
(ii) PA . PB = PC . PD triangle

Solution :-
Given that : Two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle.
Prove that : (i) Δ PAC ~ Δ PDB 
            (ii) PA . PB = PC . PD
Proof : (i) In Δ PAC and Δ PDB,
        ∠P = ∠P             (Common angle)
        ∠PAC = ∠PDB         (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)
        Δ PAC ∼ Δ PDB       (By AA similarity rule)
    
Proof : (ii) We know that the corresponding sides of similar triangles are proportional.
        Therefore, PA/PD = AC/DB = PC/PB
        PA/PD = PC/PB
        PA . PB = PC . PD
    

Question-9 :-  In Figure, D is a point on side BC of Δ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC. triangle

Solution :-
Given that : D is a point on side BC of Δ ABC such that BD/CD = AB/AC.
Prove that : AD is the bisector of ∠BAC.
Construction : Let us extend BA to P such that AP = AC. Join PC.
    triangle
Proof : It is given that BD/CD = AB/AC
        BD/CD = AP/AC           (AP = AC)
        By using the converse of basic proportionality theorem, we obtain 
        AD || PC
        ∠BAD = ∠APC             (Corresponding angles) ....... (i)
        ∠DAC = ∠ACP             (Alternate interior angles) ....... (ii)
        By construction, we have
        AP = AC
        ∠APC = ∠ACP ....... (iii)
        On comparing equations (i), (ii), and (iii), we obtain 
        ∠BAD = ∠APC
        AD is the bisector of the angle BAC.
    

Question-10 :-  Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds? triangle

Solution :-
    triangle
    Let AB be the height of the tip of the fishing rod from the water surface. Let BC be
    the horizontal distance of the fly from the tip of the fishing rod.
    Then, AC is the length of the string.
    AC can be found by applying Pythagoras theorem in ΔABC. 
    AC2 = AB2 + BC2
    AB2 = 1.82 + 2.42 
    AB2 = 3.24 + 5.76 
    AB2 = 9
    AB = 3
    Thus, the length of the string out is 3 m.
    She pulls the string at the rate of 5 cm per second.
    Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m
    triangle
    Let the fly be at point D after 12 seconds.
    Length of string out after 12 seconds is AD.
    AD = AC − String pulled by Nazima in 12 seconds
    AD = 3.00 − 0.6
    AD = 2.4
    In Δ ADB, using pythagoras theorem
    AB2 + BD2 = AD2
    1.82 + BD2 = 2.42
    BD2 = 5.76 − 3.24 
    BD2 = 2.52 
    BD = 1.587
    Horizontal distance of fly = BD + 1.2 = 1.587 + 1.2 = 2.787 = 2.79 m
    
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