TOPICS

Exercise - 6.6

Triangles

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

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. 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) DM^{2} = DN . MC

(ii) DN^{2} = DM . AN

Given that : In Δ ABC, D is a point on hypotenuse AC and BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that : (i) DM^{2}= DN . MC (ii) DN^{2}= DM . AN Construction : Let us join DB. Proof : (i) We have, DN || CB, DM || AB, and ∠B = 90^{o}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 = 90^{o}∠2 + ∠3 = 90^{o}........ (i) In Δ CDM, ∠1 + ∠2 + ∠DMC = 180^{o}∠1 + ∠2 = 90^{o}........ (ii) In Δ DMB, ∠3 + ∠DMB + ∠4 = 180^{o}∠3 + ∠4 = 90^{o}........ (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) DM^{2}= DN . MC

Proof : (ii) ) In right Δ DBN, ∠5 + ∠7 = 90^{o}........ (iv) In right Δ DAN, ∠6 + ∠8 = 90^{o}........ (v) D is the foot of the perpendicular drawn from B to AC. Therefore, ∠ADB = 90^{o}∠5 + ∠6 = 90^{o}........ (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) DN^{2}= AN . DM

**Question-3 :-** In Figure, ABC is a triangle in which ∠ ABC > 90^{o} and AD ⊥ CB produced. Prove that AC^{2} = AB^{2} + BC^{2} + 2 BC . BD.

Given that : ABC is a triangle in which ∠ ABC > 90^{o}and AD ⊥ CB produced. Prove that : AC^{2}= AB^{2}+ BC^{2}+ 2 BC . BD Proof : Applying Pythagoras theorem in Δ ADB, we obtain AB^{2}= AD^{2}+ DB^{2}....... (i) Applying Pythagoras theorem in Δ ACD, we obtain AC^{2}= AD^{2}+ DC^{2}AC^{2}= AD^{2}+ (DB + BC)^{2}AC^{2}= AD^{2}+ DB^{2}+ BC^{2}+ 2DB . BC By using equation (i), we get AC^{2}= AB^{2}+ BC^{2}+ 2DB . BC

**Question-4 :-** In Figure, ABC is a triangle in which ∠ ABC < 90^{o} and AD ⊥ BC. Prove that AC^{2} = AB^{2} + BC^{2} – 2 BC . BD.

Given that : ABC is a triangle in which ∠ ABC < 90^{o}and AD ⊥ BC. Prove that : AC^{2}= AB^{2}+ BC^{2}– 2 BC . BD Proof : Applying Pythagoras theorem in Δ ADB, we obtain AD^{2}+ DB^{2}= AB^{2}AD^{2}= AB^{2}− DB^{2}........ (i) Applying Pythagoras theorem in Δ ADC, we obtain AD^{2}+ DC^{2}= AC^{2}By using eqaution (i), we get AB^{2}− BD^{2}+ DC^{2}= AC^{2}AB^{2}− BD^{2}+ (BC − BD)^{2}= AC^{2}AC^{2}= AB^{2}− BD^{2}+ BC^{2}+ BD^{2}− 2 BC . BD AC^{2}= AB^{2}+ BC^{2}− 2 BC . BD

**Question-5 :-** In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC^{2} = AD^{2} + BC . DM + (BC/2)^{2}

(ii) AB^{2} = AD^{2} - BC . DM + (BC/2)^{2}

(iii) AC^{2} + AB^{2} = 2AD^{2} + BC^{2}/2

Given that : AD is a median of a Δ ABC and AM ⊥ BC. Prove that : (i) AC^{2}= AD^{2}+ BC . DM + (BC/2)^{2}(ii) AB^{2}= AD^{2}- BC . DM + (BC/2)^{2}(iii) AC^{2}+ AB^{2}= 2AD^{2}+ BC^{2}/2 Proof : (i) Applying Pythagoras theorem in Δ AMD, we obtain AM^{2}+ MD^{2}= AD^{2}.......... (i) Applying Pythagoras theorem in Δ AMC, we obtain AM^{2}+ MC^{2}= AC^{2}AM^{2}+ (MD + DC)^{2}= AC^{2}(AM^{2}+ MD^{2}) + DC^{2}+ 2 MD . DC = AC^{2}By using equation (i), we get AD^{2}+ DC^{2}+ 2 MD . DC = AC^{2}Using the result, DC = BC/2 , we obtain AD^{2}+ (BC/2)^{2}+ 2 MD(BC/2) = AC^{2}AD^{2}+ (BC/2)^{2}+ MD . BC = AC^{2}

Proof : (ii) Applying Pythagoras theorem in Δ ABM, we obtain AB^{2}= AM^{2}+ MB^{2}AB^{2}= (AD^{2}− DM^{2}) + MB^{2}AB^{2}= (AD^{2}− DM^{2}) + (BD − MD)^{2}AB^{2}= AD^{2}− DM^{2}+ BD^{2}+ MD^{2}− 2 BD . MD AB^{2}= AD^{2}+ BD^{2}− 2 BD . MD AB^{2}= AD^{2}+ (BC/2)^{2}- 2 x BC/2 x MD AB^{2}= AD^{2}+ (BC/2)^{2}- BC . MD

Proof : (iii) Applying Pythagoras theorem in Δ ABM, we obtain AM^{2}+ MB^{2}= AB^{2}....... (i) Applying Pythagoras theorem in Δ AMC, we obtain AM^{2}+ MC^{2}= AC^{2}....... (ii) Adding equations (i) and (ii), we obtain 2AM^{2}+ MB^{2}+ MC^{2}= AB^{2}+ AC^{2}2AM^{2}+ (BD − DM)^{2}+ (MD + DC)^{2}= AB^{2}+ AC^{2}2AM^{2}+ BD^{2}+ DM^{2}− 2 BD . DM + MD^{2}+ DC^{2}+ 2 MD . DC = AB^{2}+ AC^{2}2AM^{2}+ 2MD^{2}+ BD^{2}+ DC^{2}+ 2MD (− BD + DC) = AB^{2}+ AC^{2}2(AM^{2}+ MD^{2}) + (BC/2)^{2}+ (BC/2)^{2}+ 2MD(-BC/2 + BC/2) = AB^{2}+ AC^{2}2AD^{2}+ BC^{2}/2 = AB^{2}+ AC^{2}

**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.

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 DE^{2}+ EA^{2}= DA^{2}........ (i) Applying Pythagoras theorem in Δ DEB, we obtain DE^{2}+ EB^{2}= DB^{2}DE^{2}+ (EA + AB)^{2}= DB^{2}(DE^{2}+ EA^{2}) + AB^{2}+ 2 EA . AB = DB^{2}DA^{2}+ AB^{2}+ 2 EA . AB = DB^{2}......... (ii) Applying Pythagoras theorem in Δ ADF, we obtain AD^{2}= AF^{2}+ FD^{2}Applying Pythagoras theorem in Δ AFC, we obtain AC^{2}= AF^{2}+ FC^{2}AC^{2}= AF^{2}+ (DC − FD)^{2}AC^{2}= AF^{2}+ DC^{2}+ FD^{2}− 2 DC . FD AC^{2}= (AF^{2}+ FD^{2}) + DC^{2}− 2 DC . FD AC^{2}= AD^{2}+ DC^{2}− 2 DC . FD ........ (iii) Since ABCD is a parallelogram, AB = CD ........ (iv) BC = AD ........ (v) In Δ DEA and Δ ADF, ∠DEA = ∠AFD (Both 90^{o}) ∠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 DA^{2}+ AB^{2}+ 2 EA . AB + AD^{2}+ DC^{2}− 2 DC . FD = DB^{2}+ AC^{2}DA^{2}+ AB^{2}+ AD^{2}+ DC^{2}+ 2 EA . AB − 2 DC . FD = DB^{2}+ AC^{2}BC^{2}+ AB^{2}+ AD^{2}+ DC^{2}+ 2 EA . AB − 2 AB . EA = DB^{2}+ AC^{2}By Using equations (iv) and (vi), we get AB^{2}+ BC^{2}+ CD^{2}+ DA^{2}= AC^{2}+ BD^{2}

**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

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. 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

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.

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. 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?

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. AC^{2}= AB^{2}+ BC^{2}AB^{2}= 1.8^{2}+ 2.4^{2}AB^{2}= 3.24 + 5.76 AB^{2}= 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 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 AB^{2}+ BD^{2}= AD^{2}1.8^{2}+ BD^{2}= 2.4^{2}BD^{2}= 5.76 − 3.24 BD^{2}= 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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