Chapter 6: Triangles

Answer:
(i) similar
(ii) similar
(iii) equilateral
(iv) (a) equal    (b) proportional

Answer:
(i) Similar figures: (a) Two equilateral triangles of different sides, (b) Two squares of different sides.
(ii) Non-similar figures: (a) A circle and a square, (b) A right triangle and an equilateral triangle.

Answer: Not similar. Corresponding angles are equal (all 90°), but corresponding sides are not in the same ratio (3/1.5 = 2, but 3/3 = 1).

Answer:
(i) EC = (AE × DB)/AD = (1×3)/1.5 = 2 cm
(ii) AD = (AB × AE)/AC = (7.2×1.8)/5.4 = 2.4 cm

Answer:
(i) PE/EQ = 3.9/3 = 1.3, PF/FR = 3.6/2.4 = 1.5 → Not equal → EF not parallel
(ii) PE/EQ = 4/4.5 = 8/9, PF/RF = 8/9 → Equal → EF || QR
(iii) PE/EQ = 0.18/(1.28-0.18)=0.18/1.1≈0.1636, PF/FR = 0.36/(2.56-0.36)=0.36/2.2≈0.1636 → Equal → EF || QR

Q3: LM || CB → AM/AB = AL/AC. LN || CD → AN/AD = AL/AC. Hence AM/AB = AN/AD.
Q4: DF || AE → BF/FE = BD/DA. DE || AC → BE/EC = BD/DA. Hence BF/FE = BE/EC.
Q5: Using BPT twice: in ΔPOQ, DE||OQ → PD/DO = PE/EQ. In ΔPOR, DF||OR → PD/DO = PF/FR. Thus PE/EQ = PF/FR → EF || QR.
Q6: Similarly using BPT in ΔPOQ and ΔPOR → BC || QR.
Q7: Line through mid-point of AB parallel to BC will divide AC in ratio 1:1 (by BPT) → bisects AC.
Q8: Using converse of BPT (Theorem 6.2), line joining mid-points of two sides divides third side in equal ratio, hence parallel.
Q9: In trapezium ABCD with AB || DC, using BPT in triangles formed by diagonals, we get AO/BO = CO/DO.
Q10: If AO/BO = CO/DO, then by converse of BPT, AB || CD, so ABCD is a trapezium.

Answer (selected pairs):
(i) ΔABC ~ ΔPQR (AAA: ∠A=∠P, ∠B=∠Q, ∠C=∠R)
(ii) ΔABC ~ ΔQRP (SSS: AB/QR = BC/RP = CA/PQ = 1/2)
(iii) Not similar (sides not proportional)
(iv) ΔMNL ~ ΔQPR (SAS: MN/QP = ML/QR = 1/2, included ∠M=∠Q)
(v) Not similar (angles not equal)
(vi) Not similar (corresponding angles not equal)

Answer:
∠DOC = 180°−125° = 55° (linear pair). In ΔODC, ∠DCO = 180°−(70°+55°) = 55°. Since ΔODC∼ΔOBA, ∠OAB = ∠OCD = 55°.

Q3: Using AA similarity in ΔAOB and ΔCOD (alternate angles), we get AO/OC = BO/OD.
Q4: ∠1=∠2 → PR = PQ. Given QR/QS = QT/PR = QT/PQ → by SAS, ΔPQS ~ ΔTQR.
Q5: ∠P = ∠RTS and ∠R common → ΔRPQ ~ ΔRTS (AA).
Q6: ΔABE ≅ ΔACD → AB=AC, AD=AE. So AB/AC = AD/AE → DE || BC → ΔADE ~ ΔABC.
Q7: (i) ∠AEP = ∠CDP=90°, ∠APE = ∠CPD → ΔAEP~ΔCDP.
(ii) ∠B common, ∠ADB=∠CEB=90° → ΔABD~ΔCBE.
(iii) ΔAEP~ΔADB (AA). (iv) ΔPDC~ΔBEC (AA).
Q8: In parallelogram, ∠A = ∠C, AB||CD → ∠ABE = ∠CFB → ΔABE~ΔCFB (AA).
Q9: ∠B=∠M=90°, ∠A common → ΔABC~ΔAMP (AA). Hence CA/PA = BC/MP.
Q10-14: Various similarity proofs using angle bisector theorem and median properties.
Q15: Height of tower = (6×28)/4 = 42 m.
Q16: Using median property and similarity, AB/PQ = AD/PM.

Answer: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio: AD/DB = AE/EC.

Answer: If corresponding angles are equal, triangles are similar. If two angles of one triangle equal two angles of another (AA), third angles automatically equal.

Answer: SSS: If corresponding sides are proportional, triangles are similar. SAS: If one angle equal and sides including it proportional, triangles are similar.