CBSE Class 10 Mathematics Triangles Assignment Set 12

Exercise 6: Optional Exercise

Question 1. In the figure, PS is the bisector of ∠QPR of ΔPQR. Prove that \( \frac{QS}{SR}=\frac{PQ}{PR} \).

Answer:
Construction: Draw RT ∥ SP meeting QP produced at T.
Since PS bisects ∠QPR, ∠QPS = ∠SPR ...(i). As RT ∥ SP: ∠SPR = ∠PRT (alternate angles) ...(ii) and ∠QPS = ∠QTR (corresponding angles) ...(iii).
From (i), (ii), (iii): ∠PRT = ∠QTR, so PT = PR (sides opposite equal angles).
In ΔQTR, PS ∥ TR, so by the Basic Proportionality Theorem \( \frac{QS}{SR}=\frac{QP}{PT} \). Since PT = PR, \( \frac{QS}{SR}=\frac{QP}{PR} \). Hence proved.
In simple words: This is the angle bisector theorem.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 2. In the given figure, D is a point on the hypotenuse AC of ΔABC such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that (i) DM² = DN × MC (ii) DN² = DM × AN.

Answer:
Since BD ⊥ AC, DM ⊥ BC and DN ⊥ AB with ∠B = 90°, BMDN is a rectangle, so DN = MB and DM = NB.
(i) In ΔBMD and ΔDMC: ∠BMD = ∠DMC = 90°. Also ∠2 + ∠3 = 90° (BD ⊥ AC) and ∠3 + ∠4 = 90° (ΔBDM), so ∠2 = ∠4. Hence ΔBMD ~ ΔDMC (AA) and \( \frac{BM}{DM}=\frac{DM}{CM} \), that is DM² = BM × MC = DN × MC.
(ii) In right ΔDBN and ΔAND: ∠BND = ∠DNA = 90° and ∠DBN = ∠ADN (both equal 90° minus ∠BDN), so ΔDBN ~ ΔADN (AA). Hence \( \frac{BN}{DN}=\frac{DN}{AN} \), so DN² = BN × AN = DM × AN (BN = DM).

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 3. In the given figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC × BD.

Answer:
In right ΔADB: AB² = AD² + DB² ...(i).
In right ΔADC: AC² = AD² + DC² = AD² + (DB + BC)² = AD² + DB² + BC² + 2DB × BC.
Using (i): AC² = AB² + BC² + 2BC × BD.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 4. In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² − 2BC × BD.

Answer:
In right ΔADB: AD² = AB² − DB² ...(i). In right ΔADC: AC² = AD² + DC².
Substituting (i) and DC = BC − BD: AC² = AB² − BD² + (BC − BD)² = AB² − BD² + BC² + BD² − 2BC × BD.
Hence AC² = AB² + BC² − 2BC × BD.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 5. In the figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that
(i) \( AC^2=AD^2+BC\cdot DM+\left(\frac{BC}{2}\right)^2 \)
(ii) \( AB^2=AD^2-BC\cdot DM+\left(\frac{BC}{2}\right)^2 \)
(iii) \( AC^2+AB^2=2AD^2+\tfrac12 BC^2 \)

Answer:
Since AD is a median, BD = DC = BC/2.
(i) In ΔAMC: AC² = AM² + MC² = AM² + (MD + DC)² = (AM² + MD²) + DC² + 2MD × DC. As AM² + MD² = AD² (ΔAMD) and DC = BC/2: \( AC^2=AD^2+BC\cdot DM+\left(\frac{BC}{2}\right)^2 \).
(ii) In ΔABM: AB² = AM² + BM² = AM² + (BD − MD)² = (AM² + MD²) + BD² − 2BD × MD = \( AD^2-BC\cdot DM+\left(\frac{BC}{2}\right)^2 \).
(iii) Adding (i) and (ii): \( AB^2+AC^2=2AD^2+2\cdot\frac{BC^2}{4}=2AD^2+\tfrac12BC^2 \).
In simple words: This is Apollonius' theorem for the median.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

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

Answer:
Let ABCD be a parallelogram. Draw DE ⊥ AB produced and AF ⊥ DC (so AF ⊥ DC at F).
In right ΔDEA: DE² + EA² = DA² ...(i). In right ΔDEB: DB² = DE² + EB² = DE² + (EA + AB)² = (DE² + EA²) + AB² + 2EA × AB, so DB² = DA² + AB² + 2EA × AB ...(ii).
In right ΔADF: AD² = AF² + FD² ...(iii). In right ΔACF: AC² = AF² + FC² = AF² + (DC − FD)² = (AF² + FD²) + DC² − 2DC × FD, so AC² = AD² + DC² − 2DC × FD ...(iv).
ΔEAD ≅ ΔFDA (AAS: ∠E = ∠F = 90°, ∠EAD = ∠ADF as alternate angles, AD common), so EA = DF. Also AB = DC. Hence 2EA × AB = 2DC × FD.
Adding (ii) and (iv): AC² + BD² = AB² + BC² + CD² + DA². Hence proved.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

Question 7. In the given figure, two chords AB and CD intersect each other at the point P. Prove that (i) ΔAPC ~ ΔDPB (ii) AP × PB = CP × DP.

Answer:
(i) In ΔAPC and ΔDPB: ∠APC = ∠DPB (vertically opposite angles) and ∠CAP = ∠BDP (angles in the same segment of chord CB). So ΔAPC ~ ΔDPB (AA).
(ii) Corresponding sides are proportional: \( \frac{AP}{DP}=\frac{CP}{BP} \), so AP × PB = CP × DP.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 8. In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA × PB = PC × PD.

Answer:
(i) ACDB is a cyclic quadrilateral, so ∠BAC + ∠PDB = 180° ...(i). Since PAB is a straight line, ∠PAC + ∠BAC = 180° ...(ii). From (i) and (ii), ∠PDB = ∠PAC.
In ΔPAC and ΔPDB: ∠P is common and ∠PAC = ∠PDB, so ΔPAC ~ ΔPDB (AA).
(ii) \( \frac{PA}{PD}=\frac{PC}{PB} \), so PA × PB = PC × PD.

Teacher's Note:
1. Write the similarity statement with corresponding vertices in the same order, then use the ratio of areas equal to the square of the ratio of corresponding sides.
2. Give a reason (AA, SAS or SSS) for every similarity you claim.

Question 9. In the given figure, D is a point on side BC of ΔABC such that \( \frac{BD}{CD}=\frac{AB}{AC} \). Prove that AD is the bisector of ∠BAC.

Answer:
Construction: Produce BA to P such that AP = AC, and join PC.
Given \( \frac{BD}{CD}=\frac{AB}{AC}=\frac{AB}{AP} \). In ΔPBC this gives AD ∥ PC (converse of the Basic Proportionality Theorem).
Then ∠BAD = ∠APC (corresponding angles) and ∠DAC = ∠ACP (alternate angles). Since AP = AC, ∠APC = ∠ACP.
Hence ∠BAD = ∠DAC, so AD bisects ∠BAC.

Teacher's Note:
1. Draw the perpendicular as stated in the construction and apply the Pythagoras theorem in both right triangles.
2. Subtract or add the two relations so that the common term cancels.

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? 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?

Answer:
In right ΔABC (B is the point directly below the tip A): AC² = AB² + BC² = 1.8² + 2.4² = 9.00, so AC = 3 m of string is out.
In 12 seconds she pulls in 12 × 5 = 60 cm = 0.6 m, so the string left is 3 − 0.6 = 2.4 m = AD.
In right ΔABD: BD² = AD² − AB² = 2.4² − 1.8² = 5.76 − 3.24 = 2.52, so BD = 1.587 m.
The fly is 3.6 − 2.4 = 1.2 m beyond B on the other side, as the figure shows she stands 1.2 m from B. Horizontal distance of the fly from her = BD + 1.2 = 2.787 m (about 2.79 m).

Teacher's Note:
1. Identify the hypotenuse (the longest side, opposite the right angle) before writing the Pythagoras relation.
2. In proofs, name the right-angled triangle in which you apply the theorem and write each relation as a numbered line.