(i) All circles are similar. Reason: All circles have same shape but different sizes.

(ii) All squares are similar. Reason: All squares have same shape (all angles 90°) but different side lengths.

(iii) All equilateral triangles are similar. Reason: All equilateral triangles have all angles 60°.

(iv) (a) equal, (b) proportional. Reason: For similarity, angles must be equal and sides proportional.

(i) Similar figures examples: All circles, all squares, all equilateral triangles.

(ii) Non-similar figures examples: A circle and a square, a triangle and a rectangle.

Both are squares (all angles 90°). Ratio of sides = 1.5/3 = 1/2 (constant).

✅ Yes, they are similar.

Part (i): By BPT: AD/DB = AE/EC → 1.5/3 = 1/EC → 1/2 = 1/EC → EC = 2 cm

Part (ii): AD/AB = AE/AC → AD/7.2 = 4.8/6.4 → AD/7.2 = 3/4 → AD = 7.2 × 3/4 = 5.4 cm

✅ EC = 2 cm, AD = 5.4 cm

(i) PE/EQ = 3.9/3 = 1.3, PF/FR = 3.6/2.4 = 1.5 → Not equal → ❌ EF ∦ QR

(ii) PE/EQ = 4/4.5 = 8/9, PF/FR = 8/9 → Equal → ✅ EF || QR

(iii) PE/EQ = 0.18/(1.28-0.18)=0.18/1.1=9/55, PF/FR=0.36/(2.56-0.36)=0.36/2.2=9/55 → Equal → ✅ EF || QR

In ΔABC, LM || CB → AM/AB = AL/AC ...(1)

In ΔADC, LN || CD → AN/AD = AL/AC ...(2)

From (1) and (2): AM/AB = AN/AD

✅ Hence proved.

In ΔABC, DE || AC → BD/DA = BE/EC ...(1)

In ΔABE, DF || AE → BF/FE = BD/DA ...(2)

From (1) and (2): BF/FE = BE/EC

✅ Hence proved.

In ΔOPQ, DE || OQ → PD/DO = PE/EQ ...(1)

In ΔOPR, DF || OR → PD/DO = PF/FR ...(2)

From (1) and (2): PE/EQ = PF/FR → By converse of BPT, EF || QR

✅ Hence proved.

In ΔOPQ, AB || PQ → OA/AP = OB/BQ ...(1)

In ΔOPR, AC || PR → OA/AP = OC/CR ...(2)

From (1) and (2): OB/BQ = OC/CR → By converse of BPT, BC || QR

✅ Hence proved.

In ΔABC, D is mid-point of AB, DE || BC.

By BPT: AD/DB = AE/EC → 1 = AE/EC → AE = EC

✅ Hence E is mid-point of AC.

In ΔABC, D and E are mid-points of AB and AC.

AD/DB = 1, AE/EC = 1 → AD/DB = AE/EC

By converse of BPT, DE || BC

✅ Hence proved.

In ΔAOB and ΔCOD: ∠OAB = ∠OCD (alternate), ∠OBA = ∠ODC (alternate), ∠AOB = ∠COD (vertically opposite)

∴ ΔAOB ~ ΔCOD → AO/CO = BO/DO → AO/BO = CO/DO

✅ Hence proved.

(i) ΔABC ~ ΔPQR (AAA criterion) - all angles equal (60°,80°,40°)

(ii) ΔABC ~ ΔPQR (SSS criterion) - sides: 3/6=4/8=5/10=1/2

(iii) ΔLMN ~ ΔPQR (SAS criterion) - LM/PQ=4/6=2/3, LN/PR=5/7.5=2/3, ∠L=∠P=70°

(iv) ΔABC ~ ΔPQR (AAA criterion) - all angles equal (50°,60°,70°)

∠DOC = 180° - 125° = 55° (linear pair)

In ΔODC: ∠DCO = 180° - (55° + 70°) = 55°

Since ΔODC ~ ΔOBA, ∠OAB = ∠DCO = 55°

✅ ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°

∠OAB = ∠OCD, ∠OBA = ∠ODC (alternate interior angles, AB || DC)

∠AOB = ∠COD (vertically opposite)

∴ ΔAOB ~ ΔCOD (AAA criterion) → OA/OC = OB/OD

✅ Hence proved.

Given QR/QS = QT/PR → QS/QR = PR/QT

∠PQS = ∠TQR (common angle)

By SAS criterion, ΔPQS ~ ΔTQR

✅ Hence proved.

In ΔRPQ and ΔRTS: ∠P = ∠RTS (given), ∠R is common

By AA criterion, ΔRPQ ~ ΔRTS

✅ Hence proved.

ΔABE ≅ ΔACD → AB = AC, AE = AD, ∠BAE = ∠CAD

In ΔADE and ΔABC: AD/AB = AE/AC, ∠DAE is common

By SAS criterion, ΔADE ~ ΔABC

✅ Hence proved.

(i) ∠AEP = ∠CDP = 90°, ∠APE = ∠CPD → ΔAEP ~ ΔCDP (AA)

(ii) ∠ADB = ∠CEB = 90°, ∠B common → ΔABD ~ ΔCBE (AA)

(iii) ∠AEP = ∠ADB = 90°, ∠A common → ΔAEP ~ ΔADB (AA)

(iv) ∠PDC = ∠BEC = 90°, ∠PCD = ∠BCE → ΔPDC ~ ΔBEC (AA)

In parallelogram ABCD, AB || CD → ∠ABE = ∠CFB (alternate)

∠BAE = ∠FCB (corresponding angles, AD || BC)

By AA criterion, ΔABE ~ ΔCFB

✅ Hence proved.

(i) ∠ABC = ∠AMP = 90°, ∠A common → ΔABC ~ ΔAMP (AA)

(ii) Since ΔABC ~ ΔAMP, corresponding sides are proportional: CA/PA = BC/MP

✅ Hence proved.

(i) ΔABC ~ ΔFEG → ∠ACB = ∠FGE. CD and GH are bisectors → ∠ACD = ∠FGH. Also ∠A = ∠F → ΔACD ~ ΔFGH → CD/GH = AC/FG

(ii) ∠DCB = ∠HGE (bisectors), ∠B = ∠E → ΔDCB ~ ΔHGE (AA)

(iii) ∠DCA = ∠HGF (bisectors), ∠A = ∠F → ΔDCA ~ ΔHGF (AA)

In isosceles ΔABC, AB=AC → ∠B = ∠C

AD⊥BC → ∠ADB = 90°, EF⊥AC → ∠EFC = 90°

Thus ∠ADB = ∠EFC and ∠ABD = ∠ECF → ΔABD ~ ΔECF (AA)

✅ Hence proved.

Given AB/PQ = BC/QR = AD/PM

Since AD and PM are medians, BD = BC/2, QM = QR/2

∴ BD/QM = (BC/2)/(QR/2) = BC/QR = AB/PQ

In ΔABD and ΔPQM: AB/PQ = BD/QM = AD/PM → ΔABD ~ ΔPQM (SSS) → ∠B = ∠Q

In ΔABC and ΔPQR: AB/PQ = BC/QR and ∠B = ∠Q → ΔABC ~ ΔPQR (SAS)

In ΔABC and ΔDAC: ∠BAC = ∠ADC (given), ∠C is common

∴ ΔABC ~ ΔDAC (AA) → CA/CD = CB/CA

Cross multiply: CA² = CB × CD

✅ Hence proved.

Given AB/PQ = AC/PR = AD/PM

In ΔABD and ΔPQM: AB/PQ = AD/PM and BD = BC/2, QM = QR/2

ΔABD ~ ΔPQM (SAS) → ∠B = ∠Q

In ΔABC and ΔPQR: AB/PQ = AC/PR and ∠B = ∠Q → ΔABC ~ ΔPQR (SAS)

By similarity: height of pole / shadow of pole = height of tower / shadow of tower

6/4 = h/28 → h = (6 × 28)/4 = 168/4 = 42

✅ Height of tower = 42 m

Given AB/PQ = AD/PM. Since AD and PM are medians, BD = BC/2, QM = QR/2

In ΔABD and ΔPQM: AB/PQ = AD/PM, ∠ADB = ∠PMQ (medians)

ΔABD ~ ΔPQM → ∠B = ∠Q

In ΔABC and ΔPQR: AB/PQ = BC/QR = 2BD/2QM = BD/QM and ∠B = ∠Q

By SAS criterion, ΔABC ~ ΔPQR