Key Concepts & Formulae at a Glance
- Similar polygons:Same number of sides, corresponding angles equal, and corresponding sides in the same ratio (proportion).
- Basic Proportionality Theorem (Thales Theorem):A line parallel to one side of a triangle, intersecting the other two sides, divides those two sides in the same ratio.
- Converse of BPT:If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
- AAA / AA similarity:If corresponding angles of two triangles are equal (all three, or just two — the third follows automatically), the triangles are similar.
- SSS similarity:If corresponding sides of two triangles are all in the same ratio, the triangles are similar.
- SAS similarity:If one angle of a triangle equals one angle of another, and the sides including that angle are in the same ratio, the triangles are similar.
Exercise 6.1
3 Questions
1Fill in the blanks using the correct word given in brackets:
(i) All circles are ___. (congruent, similar)
(ii) All squares are ___. (similar, congruent)
(iii) All ___ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ___ and (b) their corresponding sides are ___. (equal, proportional)
(i) All circles are similar — same shape always, but radii can differ.
(ii) All squares are similar — all angles are 90° and all sides are in the same ratio, regardless of size.
(iii) All equilateral triangles are similar — every equilateral triangle has three 60° angles, so any two are automatically equiangular.
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.
(ii) All squares are similar — all angles are 90° and all sides are in the same ratio, regardless of size.
(iii) All equilateral triangles are similar — every equilateral triangle has three 60° angles, so any two are automatically equiangular.
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.
2Give two different examples of pair of (i) similar figures (ii) non-similar figures.
(i) Similar figures
• Any two squares — both have four 90° angles and four equal sides, so corresponding angles are equal and corresponding sides are always in the same ratio.
• Any two equilateral triangles — both have three 60° angles, so any pair is automatically similar regardless of side length.
• Any two equilateral triangles — both have three 60° angles, so any pair is automatically similar regardless of side length.
(ii) Non-similar figures
• A square and a rectangle (that isn't a square) — corresponding angles are equal (all 90°), but corresponding sides are not in the same ratio.
• An equilateral triangle and a right-angled triangle — corresponding angles are not equal, so they can never be similar.
• An equilateral triangle and a right-angled triangle — corresponding angles are not equal, so they can never be similar.
3State whether the following quadrilaterals are similar or not (square ABCD of side 3 cm; rhombus PQRS of side 1.5 cm, drawn tilted):
Square ABCD (all angles 90°) and rhombus PQRS (sheared, angles ≠ 90°) — schematic, angles exaggerated for clarity.
Corresponding sides: ABPQ = BCQR = CDRS = DASP = 31.5 = 2 — so the sides are in the same ratio.
But ABCD is a square, so every interior angle is 90°. PQRS is a rhombus that has been sheared — its interior angles are not 90° (they are, say, 60° and 120° as drawn).
Corresponding angles are therefore not equal, even though corresponding sides are proportional.
But ABCD is a square, so every interior angle is 90°. PQRS is a rhombus that has been sheared — its interior angles are not 90° (they are, say, 60° and 120° as drawn).
Corresponding angles are therefore not equal, even though corresponding sides are proportional.
Not similar — both conditions (equal angles and proportional sides) must hold together, and here only the sides condition is satisfied.
Exercise 6.2
10 Questions
1In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
DE || BC in both triangles — schematic, drawn to the given ratios (not exact NCERT figure).
(i) AD = 1.5 cm, DB = 3 cm, AE = 1 cm
DE || BC ⇒ ADDB = AEEC (Basic Proportionality Theorem)
1.53 = 1EC
EC = 1 × 31.5
1.53 = 1EC
EC = 1 × 31.5
EC = 2 cm
(ii) DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm
DE || BC ⇒ ADDB = AEEC (BPT)
AD7.2 = 1.85.4
AD = 7.2 × 1.85.4
AD7.2 = 1.85.4
AD = 7.2 × 1.85.4
AD = 2.4 cm
2E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
By the converse of BPT, EF || QR exactly when PEEQ = PFFR.
(i) PEEQ = 3.93 = 1.3; PFFR = 3.62.4 = 1.5. Not equal.
(ii) PEQE = 44.5 = 89; PFRF = 89. Equal.
(iii) QE = PQ−PE = 1.28−0.18 = 1.10; RF = PR−PF = 2.56−0.36 = 2.20. PEQE = 0.181.10 = 955; PFRF = 0.362.20 = 955. Equal.
(i) PEEQ = 3.93 = 1.3; PFFR = 3.62.4 = 1.5. Not equal.
(ii) PEQE = 44.5 = 89; PFRF = 89. Equal.
(iii) QE = PQ−PE = 1.28−0.18 = 1.10; RF = PR−PF = 2.56−0.36 = 2.20. PEQE = 0.181.10 = 955; PFRF = 0.362.20 = 955. Equal.
(i)
EF is not parallel to QR.
(ii)
EF is parallel to QR.
(iii)
EF is parallel to QR.
3In Fig. 6.18, if LM || CB and LN || CD, prove that AMAB = ANAD.
L on AC (below, off-diagram), M on AB, N on AD, with LM || CB and LN || CD — schematic.
In ΔABC, LM || CB
So, AMAL = ABAC (BPT, taken as AMMB = ALLC, then componendo)
i.e. AMAB = ALAC ...(1)
In ΔACD, LN || CD
So, ANAL = ADAC (same reasoning)
i.e. ANAD = ALAC ...(2)
From (1) and (2), AMAB = ALAC = ANAD
So, AMAL = ABAC (BPT, taken as AMMB = ALLC, then componendo)
i.e. AMAB = ALAC ...(1)
In ΔACD, LN || CD
So, ANAL = ADAC (same reasoning)
i.e. ANAD = ALAC ...(2)
From (1) and (2), AMAB = ALAC = ANAD
AMAB = ANAD, as required.
4In Fig. 6.19, DE || AC and DF || AE. Prove that BFFE = BEEC.
D on AB, F and E on BC, with DE || AC and DF || AE — schematic.
In ΔBCA, DE || AC
So, BDDA = BEEC ...(1) (BPT)
In ΔBEA, DF || AE
So, BDDA = BFFE ...(2) (BPT)
From (1) and (2), BEEC = BDDA = BFFE
So, BDDA = BEEC ...(1) (BPT)
In ΔBEA, DF || AE
So, BDDA = BFFE ...(2) (BPT)
From (1) and (2), BEEC = BDDA = BFFE
BFFE = BEEC, as required.
5In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
D, E on PQ with O between; F, G on PR — DE || OQ and DF || OR — schematic (labelled to match the proof, not exact NCERT letters).
In ΔPOQ, DE || OQ
So, PEEQ = PDDO ...(1) (BPT)
In ΔPOR, DF || OR
So, PFFR = PDDO ...(2) (BPT)
From (1) and (2), PEEQ = PFFR
So, in ΔPQR, EF || QR (converse of BPT)
So, PEEQ = PDDO ...(1) (BPT)
In ΔPOR, DF || OR
So, PFFR = PDDO ...(2) (BPT)
From (1) and (2), PEEQ = PFFR
So, in ΔPQR, EF || QR (converse of BPT)
EF || QR, as required.
6In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
A, P on OQ and B, C on OR with AB || PQ and AC || PR — schematic.
In ΔOPQ, AB || PQ
So, OAAP = OBBQ ...(1) (BPT)
In ΔOPR, AC || PR
So, OAAP = OCCR ...(2) (BPT)
From (1) and (2), OBBQ = OCCR
So, in ΔOQR, BC || QR (converse of BPT)
So, OAAP = OBBQ ...(1) (BPT)
In ΔOPR, AC || PR
So, OAAP = OCCR ...(2) (BPT)
From (1) and (2), OBBQ = OCCR
So, in ΔOQR, BC || QR (converse of BPT)
BC || QR, as required.
7Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Let ΔABC be a triangle with D the mid-point of AB, and let a line through D parallel to BC meet AC at E.
Since D is the mid-point of AB, AD = DB, i.e. ADDB = 1.
Since DE || BC, by Theorem 6.1 (BPT): ADDB = AEEC
So, AEEC = 1, i.e. AE = EC
Since D is the mid-point of AB, AD = DB, i.e. ADDB = 1.
Since DE || BC, by Theorem 6.1 (BPT): ADDB = AEEC
So, AEEC = 1, i.e. AE = EC
E is the mid-point of AC — the line through the mid-point of one side, parallel to a second side, bisects the third side.
8Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Let ΔABC be a triangle with D the mid-point of AB and E the mid-point of AC.
Then AD = DB and AE = EC, so ADDB = 1 and AEEC = 1.
So, ADDB = AEEC
By Theorem 6.2 (converse of BPT), since D and E divide AB and AC in the same ratio, DE || BC
Then AD = DB and AE = EC, so ADDB = 1 and AEEC = 1.
So, ADDB = AEEC
By Theorem 6.2 (converse of BPT), since D and E divide AB and AC in the same ratio, DE || BC
The line joining the mid-points of two sides of a triangle is parallel to the third side.
9ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOBO = CODO.
Trapezium ABCD, AB || DC, diagonals AC and BD meet at O.
Through O, draw OE || DC (and hence OE || AB), with E on AD.
In ΔADC, OE || DC
So, AEED = AOOC ...(1) (BPT)
Alternatively (the standard method): in ΔAOB and ΔCOD,
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate angles, AB || DC)
So ΔAOB ~ ΔCOD (AA similarity)
Therefore AOCO = BODO (corresponding sides of similar triangles)
In ΔADC, OE || DC
So, AEED = AOOC ...(1) (BPT)
Alternatively (the standard method): in ΔAOB and ΔCOD,
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate angles, AB || DC)
So ΔAOB ~ ΔCOD (AA similarity)
Therefore AOCO = BODO (corresponding sides of similar triangles)
AOBO = CODO, as required.
10The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO. Show that ABCD is a trapezium.
Given: AOBO = CODO, i.e. AOCO = BODO.
In ΔAOB and ΔCOD:
AOCO = BODO (given)
∠AOB = ∠COD (vertically opposite angles)
So ΔAOB ~ ΔCOD (SAS similarity criterion)
Therefore ∠OAB = ∠OCD (corresponding angles of similar triangles)
These are alternate angles formed by AB and DC with transversal AC.
So, AB || DC
In ΔAOB and ΔCOD:
AOCO = BODO (given)
∠AOB = ∠COD (vertically opposite angles)
So ΔAOB ~ ΔCOD (SAS similarity criterion)
Therefore ∠OAB = ∠OCD (corresponding angles of similar triangles)
These are alternate angles formed by AB and DC with transversal AC.
So, AB || DC
Since AB || DC, ABCD is a trapezium.
Exercise 6.3
16 Questions
1State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used and the pairs of similar triangles in symbolic form.
All six pairs from Fig 6.34, schematic (anglessides as given, not to exact scale).
| Pair | Given | Check | Result |
|---|---|---|---|
| (i) | ΔABC: 60°,80°,40°; ΔPQR: 60°,80°,40° | All corresponding angles equal | Similar — AAA, ΔABC ~ ΔPQR |
| (ii) | ΔABC: 2, 2.5, 3; ΔPQR: 4, 5, 6 | Sorted ratio 2:4 = 2.5:5 = 3:6 = 1:2 | Similar — SSS, ΔABC ~ ΔQRP |
| (iii) | ΔLMP: 2, 2.7, 3; ΔDEF: 4, 5, 6 | Sorted ratio 2:4=0.5, 2.7:5=0.54, 3:6=0.5 — not equal | Not similar |
| (iv) | ΔMNL: MN=2.5, ∠M=70°, ML=5; ΔQPR: QP=5, ∠Q=70°, QR=10 | MNQP = MLQR = 12, included angle equal | Similar — SAS, ΔMNL ~ ΔQPR |
| (v) | ΔABC: AB=2.5, ∠A=80°, BC=3; ΔDEF: DE=5, ∠F=80°, EF=6 | The 80° angle is not the angle included between the two proportional sides | Not similar |
| (vi) | ΔDEF: ∠D=70°,∠F=80° ⇒ ∠E=30°; ΔPQR: ∠Q=80°,∠R=30° ⇒ ∠P=70° | ∠D=∠P, ∠F=∠Q, ∠E=∠R | Similar — AA, ΔDEF ~ ΔPRQ |
(v) is the key trap to notice: ABEF = 2.56 ≠ BCDE = 35, and the given 80° angle sits at A (between AB and AC) in the first triangle but at F (between DF and EF) in the second — not the same position relative to the sides we can actually compare. SAS needs the equal angle to be included between the two proportional sides, so this pair fails despite one matching angle.
2In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Two lines DC (top) and AB (bottom) with cevians DB and CA crossing at O.
D, O, B are collinear (DB is a straight line), so:
∠DOC + ∠BOC = 180° (linear pair)
∠DOC = 180° − 125° = 55°
In ΔODC, angles sum to 180°:
∠DOC + ∠ODC + ∠DCO = 180°
55° + 70° + ∠DCO = 180°
∠DCO = 55°
Since ΔODC ~ ΔOBA, corresponding angles are equal: ∠OAB corresponds to ∠OCD.
∠OAB = ∠DCO = 55°
∠DOC + ∠BOC = 180° (linear pair)
∠DOC = 180° − 125° = 55°
In ΔODC, angles sum to 180°:
∠DOC + ∠ODC + ∠DCO = 180°
55° + 70° + ∠DCO = 180°
∠DCO = 55°
Since ΔODC ~ ΔOBA, corresponding angles are equal: ∠OAB corresponds to ∠OCD.
∠OAB = ∠DCO = 55°
∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°
3Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC = OBOD.
In ΔOAB and ΔOCD:
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate interior angles, AB || DC, transversal AC)
So ΔOAB ~ ΔOCD (AA similarity criterion)
Therefore, corresponding sides are proportional: OAOC = OBOD
∠AOB = ∠COD (vertically opposite angles)
∠OAB = ∠OCD (alternate interior angles, AB || DC, transversal AC)
So ΔOAB ~ ΔOCD (AA similarity criterion)
Therefore, corresponding sides are proportional: OAOC = OBOD
OAOC = OBOD, as required.
4In Fig. 6.36, QRQS = QTPR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
S on PR with ∠1 = ∠PQS-side angle and ∠2 marked on QR — schematic for the QRQS = QTPR relation.
∠1 = ∠2 is given, and ∠1 is the angle ∠PQR (at Q, in ΔPQR) while ∠2 is the angle ∠PRQ (at R) — so ∠PQR = ∠PRQ, which makes ΔPQR isosceles with PQ = PR.
Since PQ = PR, we can rewrite the given ratio QRQS = QTPR as QRQS = QTPQ.
In ΔPQS and ΔTQR:
QRQS = QTPQ ⇒ PQQS = QTQR (rearranging)
∠PQS = ∠TQR (same angle at Q, since ∠1 = ∠2 places S, T consistently)
So ΔPQS ~ ΔTQR (SAS similarity criterion)
Since PQ = PR, we can rewrite the given ratio QRQS = QTPR as QRQS = QTPQ.
In ΔPQS and ΔTQR:
QRQS = QTPQ ⇒ PQQS = QTQR (rearranging)
∠PQS = ∠TQR (same angle at Q, since ∠1 = ∠2 places S, T consistently)
So ΔPQS ~ ΔTQR (SAS similarity criterion)
ΔPQS ~ ΔTQR, as required.
5S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
In ΔRPQ and ΔRTS:
∠RPQ = ∠RTS (given, ∠P = ∠RTS)
∠PRQ = ∠TRS (common angle at R — same angle for both triangles)
So ΔRPQ ~ ΔRTS (AA similarity criterion)
∠RPQ = ∠RTS (given, ∠P = ∠RTS)
∠PRQ = ∠TRS (common angle at R — same angle for both triangles)
So ΔRPQ ~ ΔRTS (AA similarity criterion)
ΔRPQ ~ ΔRTS, as required.
6In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
D on AC, E on AB, with BE and CD drawn — ΔABE ≅ ΔACD given.
ΔABE ≅ ΔACD (given)
So AB = AC and AE = AD (corresponding sides of congruent triangles)
Since AB = AC and AE = AD, we can write:
ADAB = AEAC (dividing AE=AD by AB=AC, both equal ratios)
In ΔADE and ΔABC:
ADAB = AEAC (shown above)
∠DAE = ∠BAC (common angle at A)
So ΔADE ~ ΔABC (SAS similarity criterion)
So AB = AC and AE = AD (corresponding sides of congruent triangles)
Since AB = AC and AE = AD, we can write:
ADAB = AEAC (dividing AE=AD by AB=AC, both equal ratios)
In ΔADE and ΔABC:
ADAB = AEAC (shown above)
∠DAE = ∠BAC (common angle at A)
So ΔADE ~ ΔABC (SAS similarity criterion)
ΔADE ~ ΔABC, as required.
7In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
AD ⊥ BC, CE ⊥ AB, meeting at P — computed so both angles at D and E are genuinely 90°.
(i) ΔAEP ~ ΔCDP
∠AEP = ∠CDP = 90° (CE ⊥ AB, AD ⊥ BC)
∠APE = ∠CPD (vertically opposite angles)
So ΔAEP ~ ΔCDP (AA similarity criterion)
∠APE = ∠CPD (vertically opposite angles)
So ΔAEP ~ ΔCDP (AA similarity criterion)
(ii) ΔABD ~ ΔCBE
∠ADB = ∠CEB = 90° (AD ⊥ BC, CE ⊥ AB)
∠ABD = ∠CBE (common angle at B)
So ΔABD ~ ΔCBE (AA similarity criterion)
∠ABD = ∠CBE (common angle at B)
So ΔABD ~ ΔCBE (AA similarity criterion)
(iii) ΔAEP ~ ΔADB
∠AEP = ∠ADB = 90° (CE ⊥ AB, AD ⊥ BC)
∠PAE = ∠DAB (common angle at A)
So ΔAEP ~ ΔADB (AA similarity criterion)
∠PAE = ∠DAB (common angle at A)
So ΔAEP ~ ΔADB (AA similarity criterion)
(iv) ΔPDC ~ ΔBEC
∠PDC = ∠BEC = 90° (AD ⊥ BC, CE ⊥ AB)
∠PCD = ∠BCE (common angle at C)
So ΔPDC ~ ΔBEC (AA similarity criterion)
∠PCD = ∠BCE (common angle at C)
So ΔPDC ~ ΔBEC (AA similarity criterion)
All four pairs are similar by the AA similarity criterion, using the right angles from the two altitudes and a shared or vertically-opposite angle in each case.
8E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
In ΔABE and ΔCFB:
∠A = ∠C (opposite angles of parallelogram ABCD are equal)
∠AEB = ∠CBF (alternate angles, AE || BC since AD || BC, transversal BE)
So ΔABE ~ ΔCFB (AA similarity criterion)
∠A = ∠C (opposite angles of parallelogram ABCD are equal)
∠AEB = ∠CBF (alternate angles, AE || BC since AD || BC, transversal BE)
So ΔABE ~ ΔCFB (AA similarity criterion)
ΔABE ~ ΔCFB, as required.
9In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ΔABC ~ ΔAMP
(ii) CAPA = BCMP
Right angle at B (ΔABC) and at M (ΔAMP), with A, C, P collinear — verified computationally.
(i) ΔABC ~ ΔAMP
∠ABC = ∠AMP = 90° (given, right angles at B and M)
∠A = ∠A (common angle)
So ΔABC ~ ΔAMP (AA similarity criterion)
∠A = ∠A (common angle)
So ΔABC ~ ΔAMP (AA similarity criterion)
(ii) CAPA = BCMP
Since ΔABC ~ ΔAMP, corresponding sides are proportional:
ABAM = BCMP = CAPA
In particular, CAPA = BCMP
ABAM = BCMP = CAPA
In particular, CAPA = BCMP
ΔABC ~ ΔAMP and CAPA = BCMP, as required.
10CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:
(i) CDGH = ACFG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Since ΔABC ~ ΔFEG: ∠A = ∠F, ∠B = ∠E, ∠ACB = ∠FGE, and ACFG = BCEG.
CD bisects ∠ACB, so ∠ACD = ∠DCB = ½∠ACB. Similarly ∠FGH = ∠HGE = ½∠FGE.
Since ∠ACB = ∠FGE, it follows that ∠ACD = ∠FGH and ∠DCB = ∠HGE.
CD bisects ∠ACB, so ∠ACD = ∠DCB = ½∠ACB. Similarly ∠FGH = ∠HGE = ½∠FGE.
Since ∠ACB = ∠FGE, it follows that ∠ACD = ∠FGH and ∠DCB = ∠HGE.
(ii) ΔDCB ~ ΔHGE
∠B = ∠E (given, ΔABC ~ ΔFEG)
∠DCB = ∠HGE (shown above)
So ΔDCB ~ ΔHGE (AA similarity criterion)
∠DCB = ∠HGE (shown above)
So ΔDCB ~ ΔHGE (AA similarity criterion)
(iii) ΔDCA ~ ΔHGF
∠A = ∠F (given, ΔABC ~ ΔFEG)
∠ACD = ∠FGH (shown above)
So ΔDCA ~ ΔHGF (AA similarity criterion)
∠ACD = ∠FGH (shown above)
So ΔDCA ~ ΔHGF (AA similarity criterion)
(i) CDGH = ACFG
From part (iii), ΔDCA ~ ΔHGF, so corresponding sides are proportional:
CDGH = ACFG = DAHF
CDGH = ACFG = DAHF
All three results follow, using the bisected angles and the given similarity ΔABC ~ ΔFEG.
11In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Isosceles ΔABC (AB=AC), E on CB produced beyond B, AD ⊥ BC, EF ⊥ AC.
Since AB = AC, ΔABC is isosceles, so ∠ABD = ∠ACD (angles opposite equal sides)
∠ACD and ∠ECF are the same angle, since D and F both lie on line BC and E, C, F use the same angle at C (∠ACB = ∠ECF, as E, B, C, D lie on the same line)
So ∠ABD = ∠ECF ...(1)
Also, ∠ADB = 90° (AD ⊥ BC) and ∠EFC = 90° (EF ⊥ AC)
So ∠ADB = ∠EFC ...(2)
From (1) and (2), ΔABD ~ ΔECF (AA similarity criterion)
∠ACD and ∠ECF are the same angle, since D and F both lie on line BC and E, C, F use the same angle at C (∠ACB = ∠ECF, as E, B, C, D lie on the same line)
So ∠ABD = ∠ECF ...(1)
Also, ∠ADB = 90° (AD ⊥ BC) and ∠EFC = 90° (EF ⊥ AC)
So ∠ADB = ∠EFC ...(2)
From (1) and (2), ΔABD ~ ΔECF (AA similarity criterion)
ΔABD ~ ΔECF, as required.
12Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig. 6.41). Show that ΔABC ~ ΔPQR.
AD is a median of ΔABC (D is the midpoint of BC); PM is a median of ΔPQR (M is the midpoint of QR).
Given: ABPQ = BCQR = ADPM.
Since D and M are the mid-points of BC and QR (AD, PM are medians), BD = ½BC and QM = ½QR.
So ABPQ = BDQM = ADPM (halving BCQR = 2BD2QM keeps the same ratio)
In ΔABD and ΔPQM:
ABPQ = BDQM = ADPM (shown above)
So ΔABD ~ ΔPQM (SSS similarity criterion)
Therefore, ∠ABD = ∠PQM, i.e. ∠ABC = ∠PQR
Now, in ΔABC and ΔPQR:
ABPQ = BCQR (given)
∠ABC = ∠PQR (shown above)
So ΔABC ~ ΔPQR (SAS similarity criterion)
Since D and M are the mid-points of BC and QR (AD, PM are medians), BD = ½BC and QM = ½QR.
So ABPQ = BDQM = ADPM (halving BCQR = 2BD2QM keeps the same ratio)
In ΔABD and ΔPQM:
ABPQ = BDQM = ADPM (shown above)
So ΔABD ~ ΔPQM (SSS similarity criterion)
Therefore, ∠ABD = ∠PQM, i.e. ∠ABC = ∠PQR
Now, in ΔABC and ΔPQR:
ABPQ = BCQR (given)
∠ABC = ∠PQR (shown above)
So ΔABC ~ ΔPQR (SAS similarity criterion)
ΔABC ~ ΔPQR, as required.
13D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
In ΔADC and ΔBAC:
∠ADC = ∠BAC (given)
∠ACD = ∠BCA (common angle at C)
So ΔADC ~ ΔBAC (AA similarity criterion)
Therefore, corresponding sides are proportional: CACB = CDCA
CA × CA = CB × CD
∠ADC = ∠BAC (given)
∠ACD = ∠BCA (common angle at C)
So ΔADC ~ ΔBAC (AA similarity criterion)
Therefore, corresponding sides are proportional: CACB = CDCA
CA × CA = CB × CD
CA² = CB × CD, as required.
14Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Given: ABPQ = ACPR = ADPM.
Produce AD to E such that AD = DE, and produce PM to N such that PM = MN. Join BE, CE, QN, RN.
Since D is the mid-point of both BC and AE, the diagonals of quadrilateral ABEC bisect each other, so ABEC is a parallelogram ⇒ CE = AB (and BE = AC). Similarly, PQNR is a parallelogram ⇒ RN = PQ (and QN = PR).
Since AD = DE and PM = MN, we get AE = 2AD and PN = 2PM, so AEPN = ADPM = ABPQ = ACPR (given ratio)
In ΔACE and ΔPRN:
ACPR = AEPN (shown above)
CERN = ABPQ = ACPR (CE=AB, RN=PQ, given ratio)
So ΔACE ~ ΔPRN (SSS similarity criterion) ⇒ ∠CAE = ∠RPN ...(∠1=∠2)
In ΔABE and ΔPQN:
ABPQ = AEPN (shown above)
BEQN = ACPR = ABPQ (BE=AC, QN=PR, given ratio)
So ΔABE ~ ΔPQN (SSS similarity criterion) ⇒ ∠BAE = ∠QPN ...(∠3=∠4)
Adding, ∠CAE+∠BAE = ∠RPN+∠QPN, i.e. ∠BAC = ∠QPR
Now in ΔABC and ΔPQR:
ABPQ = ACPR (given)
∠BAC = ∠QPR (shown above)
So ΔABC ~ ΔPQR (SAS similarity criterion)
Produce AD to E such that AD = DE, and produce PM to N such that PM = MN. Join BE, CE, QN, RN.
Since D is the mid-point of both BC and AE, the diagonals of quadrilateral ABEC bisect each other, so ABEC is a parallelogram ⇒ CE = AB (and BE = AC). Similarly, PQNR is a parallelogram ⇒ RN = PQ (and QN = PR).
Since AD = DE and PM = MN, we get AE = 2AD and PN = 2PM, so AEPN = ADPM = ABPQ = ACPR (given ratio)
In ΔACE and ΔPRN:
ACPR = AEPN (shown above)
CERN = ABPQ = ACPR (CE=AB, RN=PQ, given ratio)
So ΔACE ~ ΔPRN (SSS similarity criterion) ⇒ ∠CAE = ∠RPN ...(∠1=∠2)
In ΔABE and ΔPQN:
ABPQ = AEPN (shown above)
BEQN = ACPR = ABPQ (BE=AC, QN=PR, given ratio)
So ΔABE ~ ΔPQN (SSS similarity criterion) ⇒ ∠BAE = ∠QPN ...(∠3=∠4)
Adding, ∠CAE+∠BAE = ∠RPN+∠QPN, i.e. ∠BAC = ∠QPR
Now in ΔABC and ΔPQR:
ABPQ = ACPR (given)
∠BAC = ∠QPR (shown above)
So ΔABC ~ ΔPQR (SAS similarity criterion)
ΔABC ~ ΔPQR, as required.
15A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Pole and tower cast shadows at the same time of day — equal angle of elevation
At the same time of day, the sun's angle of elevation is the same for both the pole and the tower, so the pole and its shadow form a right triangle similar to the tower and its shadow (both right-angled at the ground, with the same angle of elevation at the tip of the shadow).
By AA similarity, heightshadow is the same ratio for both:
64 = height of tower28
height of tower = 6 × 284
By AA similarity, heightshadow is the same ratio for both:
64 = height of tower28
height of tower = 6 × 284
The height of the tower is 42 m.
16If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that ABPQ = ADPM.
Since ΔABC ~ ΔPQR: ∠B = ∠Q, and ABPQ = BCQR.
Since D and M are mid-points of BC and QR, BD = ½BC and QM = ½QR, so:
ABPQ = BCQR = 2BD2QM = BDQM
In ΔABD and ΔPQM:
ABPQ = BDQM (shown above)
∠B = ∠Q (given, ΔABC ~ ΔPQR)
So ΔABD ~ ΔPQM (SAS similarity criterion)
Therefore ABPQ = ADPM (corresponding sides of similar triangles)
Since D and M are mid-points of BC and QR, BD = ½BC and QM = ½QR, so:
ABPQ = BCQR = 2BD2QM = BDQM
In ΔABD and ΔPQM:
ABPQ = BDQM (shown above)
∠B = ∠Q (given, ΔABC ~ ΔPQR)
So ΔABD ~ ΔPQM (SAS similarity criterion)
Therefore ABPQ = ADPM (corresponding sides of similar triangles)
ABPQ = ADPM, as required.
Frequently Asked Questions
They give the same conclusion. AAA requires all three pairs of corresponding angles to be equal; AA only requires two pairs to be equal, because the third pair is automatically equal by the angle sum property of a triangle. AA is just the more efficient version of AAA.
SAS similarity needs the equal angle to be the included angle — the angle formed between the two sides being compared. If the equal angle sits at a different vertex, or the two proportional sides don't actually meet at that angle, the triangles are not guaranteed to be similar even though one angle and two sides happen to match. Exercise 6.3, Q1(v) is a worked example of exactly this trap.
The BPT (Theorem 6.1) starts from a line drawn parallel to one side of a triangle and concludes that it divides the other two sides in the same ratio. The converse (Theorem 6.2) starts from a line dividing two sides in the same ratio and concludes that the line must be parallel to the third side. One goes parallel→ratio, the other goes ratio→parallel.
The order of vertices in "ΔABC ~ ΔDEF" records exactly which angles and sides correspond — A with D, B with E, C with F. Writing "ΔABC ~ ΔEDF" for the same pair would claim a different (and usually wrong) correspondence, so equal angles must line up correctly in the notation itself — as in Q1(ii) and Q1(vi) of Exercise 6.3.
Yes — a printable Formula Card covering all 7 units, including Triangles, is available for ₹99.
Continue to the Next Chapter
Chapter 7 (Coordinate Geometry) moves from similarity into the distance and section formulas.