Chapter 6: Triangles

Solution:
Pick the correct option for each blank by recalling the definition of similarity — same shape but not necessarily the same size.

(i) similar
(ii) similar
(iii) equilateral
(iv) equal, proportional

Solution:
(i) Two everyday examples of similar figures:
  (a) A passport-size photograph and a postcard-size photograph of the same person printed from the same negative — they have the same shape but different sizes.
  (b) Any two equilateral triangles, regardless of their side lengths.

(ii) Two everyday examples of non-similar figures:
  (a) A square and a rhombus — both have four equal sides, but their angles differ, so they are not necessarily similar.
  (b) A circular dining table and a rectangular dining table — they have completely different shapes.

Solution:
The given quadrilaterals are not similar, because their corresponding angles are not equal — a necessary condition for similarity of polygons is not met here.

Solution:
Since \(DE \| BC\), the Basic Proportionality Theorem applies in both figures. Set up the ratio and solve for the unknown.

Figure (i):
\(\dfrac{\mathrm{AD}}{\mathrm{BD}}=\dfrac{\mathrm{AE}}{\mathrm{EC}} \Rightarrow \dfrac{1.5}{3}=\dfrac{1}{\mathrm{EC}} \Rightarrow \mathrm{EC}=2 \text{ cm}\)

Figure (ii):
\(\dfrac{\mathrm{AD}}{\mathrm{BD}}=\dfrac{\mathrm{AE}}{\mathrm{EC}} \Rightarrow \dfrac{\mathrm{AD}}{7.2}=\dfrac{1.8}{5.4} \Rightarrow \mathrm{AD}=2.4 \text{ cm}\)

Solution:
Check whether the ratios of divided segments are equal. If they are, use the converse of the Basic Proportionality Theorem to conclude \(EF \| QR\).

(i)
\(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{3.9}{3}=1.3\)
\(\dfrac{\mathrm{PF}}{\mathrm{FR}}=\dfrac{3.6}{2.4}=\dfrac{3}{2}=1.5\)
Since \(\dfrac{\mathrm{PE}}{\mathrm{EQ}} \neq \dfrac{\mathrm{PF}}{\mathrm{FR}}\), \(EF\) is not parallel to \(QR\).

(ii)
\(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{4}{4.5}=\dfrac{8}{9}\) and \(\dfrac{\mathrm{PF}}{\mathrm{RF}}=\dfrac{8}{9}\)
Since \(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{\mathrm{PF}}{\mathrm{RF}}\), by the converse of the Basic Proportionality Theorem, \(EF \| QR\).

(iii)
\(\dfrac{\mathrm{PQ}}{\mathrm{PE}}=\dfrac{1.28}{0.18}=\dfrac{64}{9}\) and \(\dfrac{\mathrm{PR}}{\mathrm{PF}}=\dfrac{2.56}{0.36}=\dfrac{64}{9}\)
Since \(\dfrac{\mathrm{PQ}}{\mathrm{PE}}=\dfrac{\mathrm{PR}}{\mathrm{PF}}\), \(EF \| QR\).

Solution:
Given: \(LM \| BC\) and \(LN \| CD\).
To prove: \(\dfrac{\mathrm{AM}}{\mathrm{AB}}=\dfrac{\mathrm{AN}}{\mathrm{AD}}\)

Apply the Basic Proportionality Theorem in each triangle separately, then link both results through the common ratio involving \(AL\) and \(LC\).

In \(\triangle ABC\), since \(LM \| BC\):
\(\dfrac{\mathrm{AM}}{\mathrm{MB}}=\dfrac{\mathrm{AL}}{\mathrm{LC}}\)    …(i)   [Basic Proportionality Theorem]

In \(\triangle ACD\), since \(LN \| CD\):
\(\dfrac{\mathrm{AN}}{\mathrm{ND}}=\dfrac{\mathrm{AL}}{\mathrm{LC}}\)    …(ii)   [Basic Proportionality Theorem]

From (i) and (ii):
\(\dfrac{\mathrm{AM}}{\mathrm{MB}}=\dfrac{\mathrm{AN}}{\mathrm{ND}} \Rightarrow \dfrac{\mathrm{MB}}{\mathrm{AM}}=\dfrac{\mathrm{ND}}{\mathrm{AN}}\)
\(\Rightarrow \dfrac{\mathrm{MB}}{\mathrm{AM}}+1=\dfrac{\mathrm{ND}}{\mathrm{AN}}+1\)
\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{AM}}=\dfrac{\mathrm{AD}}{\mathrm{AN}} \Rightarrow \dfrac{\mathrm{AM}}{\mathrm{AB}}=\dfrac{\mathrm{AN}}{\mathrm{AD}}\)    Hence proved.

Solution:
Apply the Basic Proportionality Theorem in each of the two triangles formed by the parallel lines, then combine the two results.

In \(\triangle ABC\), since \(DE \| AC\):
\(\dfrac{\mathrm{BE}}{\mathrm{EC}}=\dfrac{\mathrm{BD}}{\mathrm{AD}}\)    …(i)   [Basic Proportionality Theorem]

In \(\triangle ABE\), since \(DF \| AE\):
\(\dfrac{\mathrm{BD}}{\mathrm{DA}}=\dfrac{\mathrm{BF}}{\mathrm{FE}}\)    …(ii)   [Basic Proportionality Theorem]

From (i) and (ii): \(\dfrac{\mathrm{BE}}{\mathrm{EC}}=\dfrac{\mathrm{BF}}{\mathrm{FE}}\)    Hence proved.

Solution:
Use the Basic Proportionality Theorem in two separate triangles to establish equal ratios, then apply its converse in the third triangle to conclude \(EF \| QR\).

In \(\triangle PQO\), since \(DE \| OQ\):
\(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{\mathrm{PD}}{\mathrm{DO}}\)    …(i)   [Basic Proportionality Theorem]

In \(\triangle POR\), since \(DF \| OR\):
\(\dfrac{\mathrm{PD}}{\mathrm{DO}}=\dfrac{\mathrm{PF}}{\mathrm{FR}}\)    …(ii)   [Basic Proportionality Theorem]

From (i) and (ii):
\(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{\mathrm{PF}}{\mathrm{FR}}\)    …(iii)

In \(\triangle PQR\), since \(\dfrac{\mathrm{PE}}{\mathrm{EQ}}=\dfrac{\mathrm{PF}}{\mathrm{FR}}\) [from (iii)],
\(\Rightarrow EF \| QR\)    [Converse of Basic Proportionality Theorem]    Hence proved.

Solution:
Apply the Basic Proportionality Theorem in two triangles to get a common ratio, then use its converse to prove the required parallel line.

In \(\triangle PQO\), since \(AB \| PQ\):
\(\dfrac{\mathrm{PA}}{\mathrm{AO}}=\dfrac{\mathrm{QB}}{\mathrm{BO}}\)    …(i)   [Basic Proportionality Theorem]

In \(\triangle POR\), since \(AC \| PR\):
\(\dfrac{\mathrm{PA}}{\mathrm{AO}}=\dfrac{\mathrm{RC}}{\mathrm{CO}}\)    …(ii)   [Basic Proportionality Theorem]

From (i) and (ii):
\(\dfrac{\mathrm{QB}}{\mathrm{BO}}=\dfrac{\mathrm{RC}}{\mathrm{CO}}\)    …(iii)

In \(\triangle QOR\), since \(\dfrac{\mathrm{QB}}{\mathrm{BO}}=\dfrac{\mathrm{RC}}{\mathrm{CO}}\) [from (iii)],
\(\Rightarrow BC \| QR\)    [Converse of Basic Proportionality Theorem]

Solution:
Given: In \(\triangle ABC\), P is the mid-point of AB (so \(AP = PB\)) and \(PQ \| BC\).
To prove: Q is the mid-point of AC.

Since \(PQ \| BC\) inside \(\triangle ABC\), the Basic Proportionality Theorem directly gives us the ratio of the divided sides.

In \(\triangle ABC\), since \(PQ \| BC\):
\(\dfrac{\mathrm{AP}}{\mathrm{PB}}=\dfrac{\mathrm{AQ}}{\mathrm{QC}}\)    …(i)   [Basic Proportionality Theorem]

Also, \(AP = PB\)    …(ii)   [Given]

Substituting (ii) into (i):
\(\dfrac{\mathrm{AP}}{\mathrm{AP}}=\dfrac{\mathrm{AQ}}{\mathrm{QC}} \Rightarrow 1=\dfrac{\mathrm{AQ}}{\mathrm{QC}}\)
\(\Rightarrow AQ = QC \Rightarrow\) Q is the mid-point of AC.

Solution:
Given: In \(\triangle ABC\), P and Q are mid-points of sides AB and AC respectively.
To prove: \(PQ \| BC\)

Since P and Q are mid-points, the segment ratios on both sides are automatically equal — this is the key step that lets us invoke the converse theorem.

Since P and Q are mid-points of AB and AC:
\(\therefore AP = PB\) and \(AQ = QC\)
\(\Rightarrow \dfrac{\mathrm{AP}}{\mathrm{PB}}=1=\dfrac{\mathrm{AQ}}{\mathrm{QC}}\)
\(\Rightarrow \dfrac{\mathrm{AP}}{\mathrm{PB}}=\dfrac{\mathrm{AQ}}{\mathrm{QC}}\)

In \(\triangle ABC\), since \(\dfrac{\mathrm{AP}}{\mathrm{PB}}=\dfrac{\mathrm{AQ}}{\mathrm{QC}}\):
\(\Rightarrow PQ \| BC\)    [Converse of Basic Proportionality Theorem]

Solution:
Given: Trapezium \(ABCD\) with \(AB \| CD\); diagonals AC and BD intersect at O.
To prove: \(\dfrac{\mathrm{AO}}{\mathrm{BO}}=\dfrac{\mathrm{CO}}{\mathrm{DO}}\)
Construction: Draw \(OE \| AB\), meeting AD at E.

Introduce the auxiliary line to create two triangles in which the Basic Proportionality Theorem can be applied, then link both results through the common ratio involving \(DE\) and \(EA\).

In \(\triangle ADB\), since \(OE \| AB\):
\(\dfrac{\mathrm{DE}}{\mathrm{EA}}=\dfrac{\mathrm{DO}}{\mathrm{BO}}\)    …(i)   [Basic Proportionality Theorem]

Since \(AB \| OE\) and \(AB \| DC\), we get \(OE \| DC\).

In \(\triangle ADC\), since \(OE \| DC\):
\(\dfrac{\mathrm{DE}}{\mathrm{EA}}=\dfrac{\mathrm{CO}}{\mathrm{AO}}\)    …(ii)   [Basic Proportionality Theorem]

From (i) and (ii):
\(\dfrac{\mathrm{CO}}{\mathrm{AO}}=\dfrac{\mathrm{DO}}{\mathrm{BO}} \Rightarrow \dfrac{\mathrm{AO}}{\mathrm{BO}}=\dfrac{\mathrm{CO}}{\mathrm{DO}}\)    Hence proved.

Solution:
Given: In quadrilateral \(ABCD\), diagonals AC and BD intersect at O such that \(\dfrac{\mathrm{AO}}{\mathrm{BO}}=\dfrac{\mathrm{CO}}{\mathrm{DO}}\).
To prove: ABCD is a trapezium.
Construction: Draw \(OE \| AB\).

The strategy is to show that \(DC \| AB\) by tracing parallel lines through the converse of the Basic Proportionality Theorem.

In \(\triangle DAB\), since \(OE \| AB\):
\(\dfrac{\mathrm{OB}}{\mathrm{OD}}=\dfrac{\mathrm{AE}}{\mathrm{ED}}\)    …(i)   [Basic Proportionality Theorem]

Also, \(\dfrac{\mathrm{OA}}{\mathrm{OC}}=\dfrac{\mathrm{OB}}{\mathrm{OD}}\) [Given]
\(\Rightarrow \dfrac{\mathrm{OA}}{\mathrm{OC}}=\dfrac{\mathrm{AE}}{\mathrm{ED}}\)    [From (i)]

In \(\triangle ADC\), since \(\dfrac{\mathrm{OA}}{\mathrm{OC}}=\dfrac{\mathrm{AE}}{\mathrm{ED}}\):
\(\Rightarrow OE \| DC\)    …(ii)   [Converse of Basic Proportionality Theorem]

Also, \(OE \| AB\)    …(iii)   [Construction]

From (ii) and (iii): \(DC \| AB\)
\(\therefore\) Quadrilateral ABCD is a trapezium.

Solution:
Examine each pair by checking angles and side ratios, then name the appropriate similarity criterion.

(i) \(\triangle ABC \sim \triangle PQR\)    [AAA similarity]

(ii) \(\dfrac{\mathrm{AB}}{\mathrm{QR}}=\dfrac{\mathrm{BC}}{\mathrm{PR}}=\dfrac{\mathrm{AC}}{\mathrm{PQ}}=\dfrac{1}{2} \Rightarrow \triangle ABC \sim \triangle QRP\)    [SSS similarity]

(iii) Not similar — the sides are not in proportion.

(iv) Not similar — SAS condition is not satisfied.

(v) Not similar.

(vi) Similar: \(\triangle DEF \sim \triangle PQR\)    [AAA similarity]

Solution:
Use the supplementary angle at O and the angle sum of a triangle to find each missing angle step by step.

Since \(\triangle ODC \sim \triangle OBA\):
\(\angle OAB = \angle OCD\)    …(i)   [Corresponding angles of similar triangles]

\(\angle DOC = 180° – 125° = 55°\)    [Angles on a straight line]

In \(\triangle DOC\), angle sum \(= 180°\):
\(\angle OCD = 180° – (70° + 55°) = 55°\)

From (i): \(\angle OAB = 55°\)

Solution:
Identify two triangles at point O that share vertically opposite angles and alternate angles due to the parallel sides of the trapezium.

In triangles AOB and COD:
\(\angle 1 = \angle 2\)    [Alternate angles, since \(AB \| DC\)]
\(\angle 4 = \angle 3\)    [Vertically opposite angles]
\(\therefore \triangle AOB \sim \triangle COD\)    [AA similarity]
\(\therefore \dfrac{\mathrm{OA}}{\mathrm{OC}}=\dfrac{\mathrm{OB}}{\mathrm{OD}}\)    Hence proved.

Solution:
First use the given angle condition to establish that two sides of \(\triangle PQR\) are equal, then substitute to rewrite the given ratio in a form that matches the SAS similarity condition.

In \(\triangle PQR\), since \(\angle 1 = \angle 2\):
\(\Rightarrow PQ = PR\)    …(i)   [Sides opposite equal angles are equal]

Given: \(\dfrac{\mathrm{QR}}{\mathrm{QS}}=\dfrac{\mathrm{QT}}{\mathrm{PR}} \Rightarrow \dfrac{\mathrm{QR}}{\mathrm{QS}}=\dfrac{\mathrm{QT}}{\mathrm{PQ}}\)    [Using (i)]

In triangles PQS and TQR:
\(\dfrac{\mathrm{QR}}{\mathrm{QS}}=\dfrac{\mathrm{QT}}{\mathrm{PQ}}\)    [Proved above]
and \(\angle 1\) is common.
\(\therefore \triangle PQS \sim \triangle TQR\)    [SAS similarity]

Solution:
Two pairs of equal angles are enough to establish AA similarity — one is given directly and the other is common to both triangles.

In triangles RPQ and RTS:
\(\angle P = \angle RTS\)    [Given]
\(\angle R\) is common to both triangles.
\(\therefore \triangle RPQ \sim \triangle RTS\)    [AA similarity]

Solution:
Extract equal corresponding sides from the congruence, form equal ratios, then combine with the common angle to apply SAS similarity.

Since \(\triangle ABE \cong \triangle ACD\):
\(\Rightarrow AB = AC\) and \(AE = AD\)
\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{AD}}=\dfrac{\mathrm{AC}}{\mathrm{AE}}\) and \(\angle A\) is common.
\(\Rightarrow \triangle ADE \sim \triangle ABC\)    [SAS similarity]

Solution:
In each part, identify two angles that are equal — one from the altitude condition (right angles) and one either common or vertically opposite — then apply AA similarity.

(i) In \(\triangle AEP\) and \(\triangle CDP\):
\(\angle AEP = \angle CDP = 90°\)    [Altitudes]
\(\angle APE = \angle CPD\)    [Vertically opposite angles]
\(\therefore \triangle AEP \sim \triangle CDP\)    [AA similarity]

(ii) In \(\triangle ABD\) and \(\triangle CBE\):
\(\angle ADB = \angle CEB = 90°\)    [Altitudes]
\(\angle B\) is common.
\(\therefore \triangle ABD \sim \triangle CBE\)    [AA similarity]

(iii) Let \(\angle DAB = x \Rightarrow \angle PAE = x\).
In \(\triangle APE\) and \(\triangle ADB\):
\(\angle PAE = \angle DAB\)    [From above]
\(\angle AEP = \angle ADB = 90°\)    [Altitudes]
\(\therefore \triangle AEP \sim \triangle ADB\)    [AA similarity]

(iv) In \(\triangle PDC\) and \(\triangle BEC\):
\(\angle C\) is common and \(\angle CDP = \angle CEB = 90°\)    [Altitudes]
\(\therefore \triangle PDC \sim \triangle BEC\)    [AA similarity]    Hence proved.

Solution:
Use the properties of a parallelogram — opposite angles are equal and alternate angles formed by a transversal with parallel sides are equal — to identify two pairs of equal angles.

In \(\triangle ABE\) and \(\triangle CFB\):
\(\angle A = \angle C\)    [Opposite angles of a parallelogram]
\(\angle 1 = \angle 2\)    [Alternate angles, since \(AB \| DC\)]
\(\therefore \triangle ABE \sim \triangle CFB\)    [AA similarity]    Hence proved.

Solution:
Both triangles share a common angle at A and each has a right angle, making AA similarity straightforward to apply. The side ratio then follows directly from the similarity.

In \(\triangle ABC\) and \(\triangle AMP\):
\(\angle A\) is common and \(\angle ABC = \angle AMP = 90°\).

(i) \(\therefore \triangle ABC \sim \triangle AMP\)    [AA similarity]

(ii) \(\dfrac{\mathrm{CA}}{\mathrm{PA}}=\dfrac{\mathrm{BC}}{\mathrm{MP}}\)    [Corresponding sides of similar triangles]    Hence proved.

Solution:
Use the similarity of the original triangles to transfer equal angles, then halve the equal angles (since CD and GH are bisectors) to set up the required smaller similarities.

(i) Since \(\triangle ABC \sim \triangle FEG\):
\(\angle A = \angle F\) and \(\angle ACB = \angle FGE\)
\(\Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle FGE \Rightarrow \angle ACD = \angle FGH\)
   [CD and GH are bisectors of \(\angle ACB\) and \(\angle FGE\)]
\(\therefore \triangle ACD \sim \triangle FGH\)    [AA similarity]
\(\Rightarrow \dfrac{\mathrm{CD}}{\mathrm{GH}}=\dfrac{\mathrm{AC}}{\mathrm{FG}}\)

(ii) In \(\triangle DCB\) and \(\triangle HGE\):
\(\angle B = \angle E\)    [\(\because \triangle ABC \sim \triangle FEG\)]
\(\angle ACB = \angle FGE \Rightarrow \dfrac{1}{2}\angle ACB = \dfrac{1}{2}\angle FGE \Rightarrow \angle DCB = \angle HGE\)
   …(iv) [CD and GH are bisectors]
\(\therefore \triangle DCB \sim \triangle HGE\)    [AA similarity]

(iii) From result (i): \(\triangle DCA \sim \triangle HGF\).

Solution:
Since the triangle is isosceles with \(AB = AC\), the base angles are equal. Pair this with the two right angles from the given perpendiculars to apply AA similarity.

In \(\triangle ABD\) and \(\triangle ECF\):
\(\angle ADB = \angle EFC = 90°\)    [\(AD \perp BC\) and \(EF \perp AC\)]
\(\angle ABD = \angle ECF\)    [\(\because AB = AC\), so base angles are equal]
\(\therefore \triangle ABD \sim \triangle ECF\)    [AA similarity]    Hence proved.

Solution:
Replace the full base with half the base (since D and M are mid-points), establish SSS similarity for the smaller triangles, extract an equal angle, and then apply SAS similarity for the original triangles.

Given: \(\dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{BC}}{\mathrm{QR}}=\dfrac{\mathrm{AD}}{\mathrm{PM}}\)    …(i)

\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\frac{1}{2}\mathrm{BC}}{\frac{1}{2}\mathrm{QR}}=\dfrac{\mathrm{AD}}{\mathrm{PM}}\)
\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{BD}}{\mathrm{QM}}=\dfrac{\mathrm{AD}}{\mathrm{PM}}\)    …(ii)   \(\left(\because BD=\dfrac{1}{2}BC \text{ and } QM=\dfrac{1}{2}QR\right)\)

\(\therefore \triangle ABD \sim \triangle PQM\)    [SSS similarity from (ii)]
\(\Rightarrow \angle B = \angle Q\)    …(iii)

In \(\triangle ABC\) and \(\triangle PQR\):
\(\dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{BC}}{\mathrm{QR}}\) [from (i)] and \(\angle B = \angle Q\) [from (iii)]
\(\Rightarrow \triangle ABC \sim \triangle PQR\)    [SAS similarity]

Solution:
Identify AA similarity between two triangles that share a common angle and have the other angle equal by the given condition, then write the proportionality of corresponding sides and cross-multiply.

In \(\triangle ABC\) and \(\triangle ADC\):
\(\angle BAC = \angle ADC\)    [Given]
\(\angle C\) is common.
\(\therefore \triangle ABC \sim \triangle DAC\)    [AA similarity]

\(\therefore \dfrac{\mathrm{CA}}{\mathrm{CD}}=\dfrac{\mathrm{CB}}{\mathrm{CA}} \Rightarrow CA^2 = CB \cdot CD\)    Hence proved.

Solution:
Given: \(\dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{AC}}{\mathrm{PR}}=\dfrac{\mathrm{AD}}{\mathrm{PM}}\)
To prove: \(\triangle ABC \sim \triangle PQR\)
Construction: Produce AD and PM to E and N respectively such that \(AD = DE\) and \(PM = MN\). Join EC and RN.

Extend the medians to create congruent triangles, then use the congruence to establish equal sides, build SSS similarity for larger triangles, and extract the required angle to finally apply SAS similarity.

In \(\triangle ABD\) and \(\triangle CDE\):
\(AD = DE\) [Construction], \(BD = DC\) [D is mid-point of BC], \(\angle ADB = \angle CDE\) [Vertically opposite]
\(\therefore \triangle ABD \cong \triangle ECD\) [SAS] \(\Rightarrow AB = CE\)    …(i)

Similarly, \(PQ = RN\)    …(ii)

From the given condition, using (i) and (ii) and the construction:
\(\dfrac{\mathrm{CE}}{\mathrm{RN}}=\dfrac{\mathrm{AC}}{\mathrm{PR}}=\dfrac{\mathrm{AE}}{\mathrm{PN}}\)
\(\Rightarrow \triangle AEC \sim \triangle PNR\)    [SSS similarity]
\(\therefore \angle DAC = \angle MPR\)    …(iii)

Similarly: \(\angle BAD = \angle QPM\)    …(iv)

Adding (iii) and (iv): \(\angle BAD + \angle DAC = \angle QPM + \angle MPR\)
\(\Rightarrow \angle BAC = \angle QPR\)    …(v)

In \(\triangle ABC\) and \(\triangle PQR\):
\(\dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{AC}}{\mathrm{PR}}\) [Given] and \(\angle BAC = \angle QPR\) [From (v)]
\(\Rightarrow \triangle ABC \sim \triangle PQR\)    [SAS similarity]    Hence proved.

Solution:
Since the sun’s rays fall at the same angle at the same instant, the triangles formed by the pole and the tower with their shadows are similar. Set up the proportion and solve.

Let the height of the tower be \(x\) m.
Both triangles are similar since \(\angle 1 = \angle 2\) (sun’s elevation angle is the same at the same time):
\(\therefore \dfrac{6}{4}=\dfrac{x}{28} \Rightarrow x = 42 \text{ m}\)

Solution:
Extract the side ratio from the given similarity, replace full bases with half-bases (using the median mid-point property), then apply SAS similarity to the smaller triangles and read off the required ratio.

Since \(\triangle ABC \sim \triangle PQR\) [Given]:
\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{BC}}{\mathrm{QR}}=\dfrac{\mathrm{AC}}{\mathrm{PR}}\)

\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\frac{1}{2}\mathrm{BC}}{\frac{1}{2}\mathrm{QR}} \Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{BD}}{\mathrm{QM}}\)    …(i)   \(\left(\because BD=\dfrac{1}{2}BC,\ QM=\dfrac{1}{2}QR\right)\)

Also, \(\angle B = \angle Q\)    …(ii)   [\(\because \triangle ABC \sim \triangle PQR\)]

From (i) and (ii): \(\triangle ABD \sim \triangle PQM\)    [SAS similarity]
\(\Rightarrow \dfrac{\mathrm{AB}}{\mathrm{PQ}}=\dfrac{\mathrm{AD}}{\mathrm{PM}}\)    Hence proved.