Triangles

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CLASS X Mathematics ~6 marks/year Ch 6 of 14
Triangles

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Two figures are similar when they have the same shape (not necessarily the same size). All congruent figures are similar, but similar figures need not be congruent.
  • Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio.
  • The Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides in the same ratio; its converse is equally important.
  • Four shortcuts to prove similarity: AAA / AA, SSS and SAS — you never need to check all six pairs of parts.
  • Board weightage: ~6 marks/year — typically one BPT-based proof (3 marks) and one similarity-criterion proof or numerical (3 marks), plus 1-mark MCQs on ratios.
Detailed notes

1. Where this chapter sits

In Class 9 you studied congruence: two figures are congruent when they have the same shape and same size. This chapter relaxes that idea — we now study figures with the same shape but not necessarily the same size. These are called similar figures.

The idea is powerful because it lets us measure things indirectly — the height of Mount Everest, the distance to the Moon, the height of a tower from its shadow — all rest on the similarity of triangles. We will build the theory, then use it to compare and compute.

2. Similar figures and similar polygons

All circles of the same radius are congruent; circles of different radii are not congruent — but they all have the same shape, so they are similar. The same holds for all squares and all equilateral triangles: any two are similar.

So: all congruent figures are similar, but similar figures need not be congruent. A circle and a square are not similar (different shapes). Two photographs of the Taj Mahal printed at different sizes are similar.

For polygons, similarity needs two conditions together:

$$\textbf{Two polygons (same number of sides) are similar if:}\quad \text{(i) corresponding angles are equal,}\ \text{and}\ \text{(ii) corresponding sides are in the same ratio.}$$

The common ratio of corresponding sides is called the scale factor (Representative Fraction) — exactly what world maps and building blueprints use.

Both conditions are essential. A square and a rectangle have all angles equal but sides not in ratio — not similar. A square and a rhombus can have sides in ratio but angles unequal — not similar. So neither condition alone is enough for polygons.

3. Similarity of triangles — definition and notation

A triangle is a polygon, so the same definition applies:

$$\triangle ABC \sim \triangle DEF \iff \angle A=\angle D,\ \angle B=\angle E,\ \angle C=\angle F\ \text{and}\ \dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}$$

The symbol "$\sim$" means "is similar to". Triangles whose corresponding angles are equal are called equiangular; Thales observed that in equiangular triangles the ratio of corresponding sides is always the same.

Order of letters matters. Just like congruence, similarity must be written with the correct correspondence of vertices. If $A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F$, you must write $\triangle ABC\sim\triangle DEF$ — writing $\triangle ABC\sim\triangle EDF$ would be wrong.

4. Basic Proportionality Theorem (Thales) — Theorem 6.1

Theorem 6.1 (BPT): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

$$\text{In } \triangle ABC,\ \text{if } DE\parallel BC \text{ (with } D \text{ on } AB,\ E \text{ on } AC),\ \text{then}\quad \dfrac{AD}{DB}=\dfrac{AE}{EC}$$
NCERT Theorem 6.1 — full proof (area method)

1. Join $BE$ and $CD$; draw $DM\perp AC$ and $EN\perp AB$.

2. $\text{ar}(ADE)=\tfrac12\,AD\times EN$ and $\text{ar}(BDE)=\tfrac12\,DB\times EN$. Dividing:

$\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)}=\dfrac{AD}{DB}\ \ \dots(1)$

3. Also $\text{ar}(ADE)=\tfrac12\,AE\times DM$ and $\text{ar}(DEC)=\tfrac12\,EC\times DM$, so $\dfrac{\text{ar}(ADE)}{\text{ar}(DEC)}=\dfrac{AE}{EC}\ \ \dots(2)$

4. $\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $BC$ and $DE$, so $\text{ar}(BDE)=\text{ar}(DEC)\ \dots(3)$.

5. From (1), (2) and (3): $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. $\blacksquare$

5. Converse of BPT — Theorem 6.2

Theorem 6.2 (Converse of BPT): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

$$\text{In } \triangle ABC,\ \text{if } \dfrac{AD}{DB}=\dfrac{AE}{EC},\ \text{then } DE\parallel BC$$

Proof idea: Suppose $DE$ is not parallel to $BC$. Draw $DE'\parallel BC$. By Theorem 6.1, $\dfrac{AD}{DB}=\dfrac{AE'}{E'C}$. But we are given $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, so $\dfrac{AE}{EC}=\dfrac{AE'}{E'C}$. Adding $1$ to both sides forces $E$ and $E'$ to coincide — so $DE$ itself is parallel to $BC$. $\blacksquare$

These two theorems are the most exam-heavy results in the chapter: BPT to set up ratios, its converse to prove lines parallel.

6. Worked Examples on BPT (Examples 1-3)

NCERT Example 1 — prove $\dfrac{AD}{AB}=\dfrac{AE}{AC}$

Given $DE\parallel BC$ in $\triangle ABC$ (with $D$ on $AB$, $E$ on $AC$).

By BPT: $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, so $\dfrac{DB}{AD}=\dfrac{EC}{AE}$.

Add $1$ to both sides: $\dfrac{DB}{AD}+1=\dfrac{EC}{AE}+1\Rightarrow\dfrac{AB}{AD}=\dfrac{AC}{AE}$.

Taking reciprocals: $\dfrac{AD}{AB}=\dfrac{AE}{AC}$. $\blacksquare$

NCERT Example 2 — trapezium, prove $\dfrac{AE}{ED}=\dfrac{BF}{FC}$

$ABCD$ is a trapezium with $AB\parallel DC$; $E,F$ on non-parallel sides $AD,BC$ with $EF\parallel AB$. Join $AC$ to meet $EF$ at $G$.

Since $AB\parallel DC$ and $EF\parallel AB$, lines parallel to the same line give $EF\parallel DC$.

In $\triangle ADC$, $EG\parallel DC$, so by BPT: $\dfrac{AE}{ED}=\dfrac{AG}{GC}\ \dots(1)$.

In $\triangle CAB$, $GF\parallel AB$, so by BPT: $\dfrac{CG}{AG}=\dfrac{CF}{BF}$, i.e. $\dfrac{AG}{GC}=\dfrac{BF}{FC}\ \dots(2)$.

From (1) and (2): $\dfrac{AE}{ED}=\dfrac{BF}{FC}$. $\blacksquare$

NCERT Example 3 — prove $\triangle PQR$ is isosceles

Given $\dfrac{PS}{SQ}=\dfrac{PT}{TR}$ and $\angle PST=\angle PRQ$.

By the converse of BPT, $\dfrac{PS}{SQ}=\dfrac{PT}{TR}\Rightarrow ST\parallel QR$.

So $\angle PST=\angle PQR$ (corresponding angles) $\dots(1)$.

Given $\angle PST=\angle PRQ\ \dots(2)$. From (1) and (2): $\angle PQR=\angle PRQ$.

Sides opposite equal angles are equal, so $PQ=PR$ — hence $\triangle PQR$ is isosceles. $\blacksquare$

7. Criteria for similarity of triangles

Just as congruence has shortcuts (SSS, SAS, ASA), similarity has shortcuts so we never check all six parts. There are three:

Theorem 6.3 (AAA): If in two triangles corresponding angles are equal, then their corresponding sides are in the same ratio (proportion) and hence the triangles are similar.

$$\angle A=\angle D,\ \angle B=\angle E,\ \angle C=\angle F\ \Rightarrow\ \dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}\ \Rightarrow\ \triangle ABC\sim\triangle DEF$$

AA corollary: By the angle-sum property, if two angles of one triangle equal two angles of another, the third angles are automatically equal. So:

$$\angle A=\angle D\ \text{and}\ \angle B=\angle E\ \Rightarrow\ \triangle ABC\sim\triangle DEF\quad(\textbf{AA criterion})$$

Theorem 6.4 (SSS): If in two triangles the sides of one are proportional to the sides of the other, then their corresponding angles are equal and the triangles are similar.

$$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}\ \Rightarrow\ \triangle ABC\sim\triangle DEF\quad(\textbf{SSS criterion})$$

Theorem 6.5 (SAS): If one angle of a triangle equals one angle of another and the sides including these angles are proportional, then the triangles are similar.

$$\angle A=\angle D\ \text{and}\ \dfrac{AB}{DE}=\dfrac{AC}{DF}\ \Rightarrow\ \triangle ABC\sim\triangle DEF\quad(\textbf{SAS criterion})$$

Each is proved by the "cut and compare" method: cut $DP=AB$, $DQ=AC$ on the larger triangle, show $PQ\parallel EF$ using BPT, deduce $\triangle ABC\cong\triangle DPQ$, then read off the similarity. A bonus result mentioned in the chapter: the RHS similarity criterion — if hypotenuse and one side of one right triangle are proportional to those of another, the triangles are similar.

8. Worked Examples on similarity criteria (Examples 4-8)

NCERT Example 4 — prove $\triangle POQ\sim\triangle SOR$

Given $PQ\parallel RS$ with $O$ the intersection of $PS$ and $QR$.

$\angle P=\angle S$ and $\angle Q=\angle R$ (alternate angles); $\angle POQ=\angle SOR$ (vertically opposite).

By the AAA (AA) criterion, $\triangle POQ\sim\triangle SOR$. $\blacksquare$

NCERT Example 5 — find $\angle P$

$\triangle ABC$: $AB=3.8,\ BC=6,\ CA=3\sqrt3,\ \angle A=80^\circ,\ \angle B=60^\circ$. $\triangle PQR$ ($=\triangle RQP$): $RQ=7.6,\ QP=12,\ PR=6\sqrt3$.

Compare ratios: $\dfrac{AB}{RQ}=\dfrac{3.8}{7.6}=\dfrac12,\quad\dfrac{BC}{QP}=\dfrac{6}{12}=\dfrac12,\quad\dfrac{CA}{PR}=\dfrac{3\sqrt3}{6\sqrt3}=\dfrac12.$

All equal, so $\triangle ABC\sim\triangle RQP$ (SSS similarity). Hence $\angle C=\angle P$.

$\angle C=180^\circ-\angle A-\angle B=180^\circ-80^\circ-60^\circ=40^\circ$, so $\angle P=40^\circ$. $\blacksquare$

NCERT Example 6 — given $OA\cdot OB=OC\cdot OD$, show $\angle A=\angle C,\ \angle B=\angle D$

From $OA\cdot OB=OC\cdot OD$ we get $\dfrac{OA}{OC}=\dfrac{OD}{OB}\ \dots(1)$.

Also $\angle AOD=\angle COB$ (vertically opposite) $\dots(2)$.

By (1), (2) and SAS similarity, $\triangle AOD\sim\triangle COB$.

Hence $\angle A=\angle C$ and $\angle D=\angle B$ (corresponding angles of similar triangles). $\blacksquare$

NCERT Example 7 — length of a shadow (indirect measurement)

A girl $90\text{ cm}=0.9\text{ m}$ tall walks away from a lamp-post at $1.2$ m/s; the lamp is $3.6$ m high. Find her shadow length after $4$ s.

Distance walked $BD=1.2\times4=4.8$ m. Let shadow $DE=x$ m. In $\triangle ABE$ and $\triangle CDE$: $\angle B=\angle D=90^\circ$ (both vertical) and $\angle E=\angle E$ (common), so $\triangle ABE\sim\triangle CDE$ (AA).

$\dfrac{BE}{DE}=\dfrac{AB}{CD}\Rightarrow\dfrac{4.8+x}{x}=\dfrac{3.6}{0.9}=4\Rightarrow 4.8+x=4x\Rightarrow 3x=4.8\Rightarrow x=1.6.$

The shadow is $1.6$ m long. $\blacksquare$

NCERT Example 8 — medians of similar triangles

$CM,RN$ are medians of $\triangle ABC,\triangle PQR$ with $\triangle ABC\sim\triangle PQR$. Prove (i) $\triangle AMC\sim\triangle PNR$, (ii) $\dfrac{CM}{RN}=\dfrac{AB}{PQ}$, (iii) $\triangle CMB\sim\triangle RNQ$.

From similarity: $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}\ \dots(1)$ and $\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R\ \dots(2)$.

(i) Since $AB=2AM$ and $PQ=2PN$, from (1): $\dfrac{2AM}{2PN}=\dfrac{CA}{RP}\Rightarrow\dfrac{AM}{PN}=\dfrac{CA}{RP}\ \dots(3)$. With $\angle MAC=\angle NPR$ (from (2)), SAS gives $\triangle AMC\sim\triangle PNR$.

(ii) From (i): $\dfrac{CM}{RN}=\dfrac{CA}{RP}=\dfrac{AB}{PQ}$ (using (1)).

(iii) From (1) and (ii), $\dfrac{CM}{RN}=\dfrac{BC}{QR}$; also $\dfrac{CM}{RN}=\dfrac{BM}{QN}$ (halves of $AB,PQ$). So $\dfrac{CM}{RN}=\dfrac{BC}{QR}=\dfrac{BM}{QN}$, giving $\triangle CMB\sim\triangle RNQ$ (SSS). $\blacksquare$

9. NCERT Exercise 6.1 — fully solved

Q1. Fill in the blanks.

  • (i) All circles are similar.
  • (ii) All squares are similar.
  • (iii) All equilateral triangles are similar.
  • (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.

Q2. Two examples of each. (i) Similar figures: a pair of any two equilateral triangles; a 35 mm and a 45 mm photo of the same scene. (ii) Non-similar figures: a square and a rectangle; a circle and a triangle.

Q3. Are the quadrilaterals similar? A square ($3$ cm side, all $90^\circ$) and a rhombus ($1.5$ cm side, angles not $90^\circ$): corresponding sides are in ratio but corresponding angles are not equal, so they are not similar.

10. NCERT Exercise 6.2 — fully solved

Q1. $DE\parallel BC$; find the unknown (BPT).

  • (i) $AD=1.5,\ DB=3,\ AE=1$. By BPT $\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow\dfrac{1.5}{3}=\dfrac{1}{EC}\Rightarrow EC=\dfrac{3\times1}{1.5}=2$ cm.
  • (ii) $DB=7.2,\ AE=1.8,\ EC=5.4$. $\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow\dfrac{AD}{7.2}=\dfrac{1.8}{5.4}=\dfrac13\Rightarrow AD=\dfrac{7.2}{3}=2.4$ cm.

Q2. Is $EF\parallel QR$? (converse of BPT) Check whether $\dfrac{PE}{EQ}=\dfrac{PF}{FR}$.

  • (i) $\dfrac{PE}{EQ}=\dfrac{3.9}{3}=1.3$ and $\dfrac{PF}{FR}=\dfrac{3.6}{2.4}=1.5$. Unequal $\Rightarrow EF\nparallel QR$.
  • (ii) $\dfrac{PE}{EQ}=\dfrac{4}{4.5}=\dfrac{8}{9}$ and $\dfrac{PF}{FR}=\dfrac{8}{9}$. Equal $\Rightarrow EF\parallel QR$.
  • (iii) $PE=0.18,\ EQ=PQ-PE=1.28-0.18=1.10$; $PF=0.36,\ FR=PR-PF=2.56-0.36=2.20$. $\dfrac{PE}{EQ}=\dfrac{0.18}{1.10}=\dfrac{9}{55}$ and $\dfrac{PF}{FR}=\dfrac{0.36}{2.20}=\dfrac{9}{55}$. Equal $\Rightarrow EF\parallel QR$.

Q3. $LM\parallel CB$ and $LN\parallel CD$; prove $\dfrac{AM}{AB}=\dfrac{AN}{AD}$. In $\triangle ACB$, $LM\parallel CB\Rightarrow\dfrac{AM}{MB}=\dfrac{AL}{LC}$. In $\triangle ACD$, $LN\parallel CD\Rightarrow\dfrac{AN}{ND}=\dfrac{AL}{LC}$. So $\dfrac{AM}{MB}=\dfrac{AN}{ND}$; adding $1$ to reciprocals as in Example 1 gives $\dfrac{AM}{AB}=\dfrac{AN}{AD}$.

Q4. $DE\parallel AC,\ DF\parallel AE$; prove $\dfrac{BF}{FE}=\dfrac{BE}{EC}$. In $\triangle ABC$, $DE\parallel AC\Rightarrow\dfrac{BD}{DA}=\dfrac{BE}{EC}\ \dots(1)$. In $\triangle ABE$, $DF\parallel AE\Rightarrow\dfrac{BD}{DA}=\dfrac{BF}{FE}\ \dots(2)$. From (1),(2): $\dfrac{BF}{FE}=\dfrac{BE}{EC}$.

Q5. $DE\parallel OQ,\ DF\parallel OR$; show $EF\parallel QR$. In $\triangle POQ$, $DE\parallel OQ\Rightarrow\dfrac{PE}{EQ}=\dfrac{PD}{DO}$. In $\triangle POR$, $DF\parallel OR\Rightarrow\dfrac{PF}{FR}=\dfrac{PD}{DO}$. So $\dfrac{PE}{EQ}=\dfrac{PF}{FR}$, and by the converse of BPT, $EF\parallel QR$.

Q6. $AB\parallel PQ,\ AC\parallel PR$; show $BC\parallel QR$. In $\triangle OPQ$, $AB\parallel PQ\Rightarrow\dfrac{OA}{AP}=\dfrac{OB}{BQ}$. In $\triangle OPR$, $AC\parallel PR\Rightarrow\dfrac{OA}{AP}=\dfrac{OC}{CR}$. So $\dfrac{OB}{BQ}=\dfrac{OC}{CR}$, and by the converse of BPT in $\triangle OQR$, $BC\parallel QR$.

Q7. (Mid-point theorem via BPT) A line through the mid-point $D$ of $AB$ parallel to $BC$ meets $AC$ at $E$. By BPT $\dfrac{AD}{DB}=\dfrac{AE}{EC}$; since $AD=DB$, the ratio is $1$, so $AE=EC$ — $E$ is the mid-point of $AC$. Hence the line bisects the third side.

Q8. (Converse mid-point via Theorem 6.2) $D,E$ are mid-points of $AB,AC$, so $\dfrac{AD}{DB}=1=\dfrac{AE}{EC}$. By the converse of BPT, $DE\parallel BC$ — the segment joining mid-points is parallel to the third side.

Q9. Trapezium $ABCD$, $AB\parallel DC$, diagonals meet at $O$; show $\dfrac{AO}{BO}=\dfrac{CO}{DO}$. Through $O$ draw $OE\parallel AB$ meeting $AD$ at $E$. In $\triangle ADC$, $OE\parallel DC\Rightarrow\dfrac{AE}{ED}=\dfrac{AO}{OC}$; in $\triangle ABD$, $OE\parallel AB\Rightarrow\dfrac{AE}{ED}=\dfrac{BO}{OD}$. So $\dfrac{AO}{OC}=\dfrac{BO}{OD}\Rightarrow\dfrac{AO}{BO}=\dfrac{CO}{DO}$.

Q10. Diagonals of quadrilateral $ABCD$ meet at $O$ with $\dfrac{AO}{BO}=\dfrac{CO}{DO}$; show $ABCD$ is a trapezium. Rewrite as $\dfrac{AO}{OC}=\dfrac{BO}{OD}$. Through $O$ draw $OE\parallel DC$ meeting $AD$ at $E$. In $\triangle ADC$, $OE\parallel DC\Rightarrow\dfrac{AE}{ED}=\dfrac{AO}{OC}=\dfrac{BO}{OD}$. By the converse of BPT in $\triangle ABD$, $OE\parallel AB$. Since $OE\parallel DC$ and $OE\parallel AB$, we get $AB\parallel DC$ — so $ABCD$ is a trapezium.

11. NCERT Exercise 6.3 — fully solved (part A)

Q1. Which pairs are similar? State the criterion and write symbolically.

  • (i) Angles $60^\circ,80^\circ,40^\circ$ in both $\Rightarrow$ AAA: $\triangle ABC\sim\triangle PQR$.
  • (ii) Sides proportional $\dfrac{2}{4}=\dfrac{2.5}{5}=\dfrac{3}{6}=\dfrac12\Rightarrow$ SSS: $\triangle ABC\sim\triangle QRP$.
  • (iii) Ratios $\dfrac{2.7}{4}\ne\dfrac{2}{5}$ — not similar.
  • (iv) $\angle M=\angle Q=70^\circ$ with the sides including this angle proportional ($\dfrac{ML}{QR}=\dfrac{MN}{QP}=\dfrac12$) $\Rightarrow$ SAS: $\triangle MNL\sim\triangle QPR$.
  • (v) Only $\angle A=\angle F=80^\circ$ given with one pair of sides — the equal angle is not the included angle, so not similar.
  • (vi) In $\triangle DEF$: $\angle D=70^\circ,\angle E=80^\circ\Rightarrow\angle F=30^\circ$. In $\triangle PQR$: $\angle Q=80^\circ,\angle R=30^\circ\Rightarrow\angle P=70^\circ$. Two angles match $\Rightarrow$ AA: $\triangle DEF\sim\triangle PQR$.

Q2. $\triangle ODC\sim\triangle OBA$, $\angle BOC=125^\circ$, $\angle CDO=70^\circ$. Find $\angle DOC,\angle DCO,\angle OAB$. $\angle DOC=180^\circ-\angle BOC=180^\circ-125^\circ=55^\circ$ (linear pair). In $\triangle ODC$: $\angle DCO=180^\circ-70^\circ-55^\circ=55^\circ$. By similarity $\angle OAB=\angle DCO=55^\circ$.

Q3. Trapezium $ABCD$, $AB\parallel DC$, diagonals meet at $O$; show $\dfrac{OA}{OC}=\dfrac{OB}{OD}$. In $\triangle OAB$ and $\triangle OCD$: $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ (alternate angles, $AB\parallel DC$); $\angle AOB=\angle COD$ (vertically opposite). So $\triangle OAB\sim\triangle OCD$ (AAA) $\Rightarrow\dfrac{OA}{OC}=\dfrac{OB}{OD}$.

Q4. $\dfrac{QR}{QS}=\dfrac{QT}{PR}$ and $\angle1=\angle2$; show $\triangle PQS\sim\triangle TQR$. $\angle1=\angle2\Rightarrow PR=PQ$, so $\dfrac{QR}{QS}=\dfrac{QT}{QP}$. With common angle $\angle Q$ (i.e. $\angle PQS=\angle TQR$) and these including sides proportional, SAS gives $\triangle PQS\sim\triangle TQR$.

Q5. $S,T$ on $PR,QR$ with $\angle P=\angle RTS$; show $\triangle RPQ\sim\triangle RTS$. $\angle RPQ=\angle RTS$ (given) and $\angle R=\angle R$ (common). By AA, $\triangle RPQ\sim\triangle RTS$.

Q6. $\triangle ABE\cong\triangle ACD$; show $\triangle ADE\sim\triangle ABC$. Congruence gives $AB=AC$ and $AE=AD$, so $\dfrac{AD}{AB}=\dfrac{AE}{AC}$. With common angle $\angle A$, SAS gives $\triangle ADE\sim\triangle ABC$.

12. NCERT Exercise 6.3 — fully solved (part B)

Q7. Altitudes $AD,CE$ of $\triangle ABC$ meet at $P$.

  • (i) $\triangle AEP\sim\triangle CDP$: $\angle AEP=\angle CDP=90^\circ$ and $\angle APE=\angle CPD$ (vertically opposite) $\Rightarrow$ AA.
  • (ii) $\triangle ABD\sim\triangle CBE$: $\angle ADB=\angle CEB=90^\circ$ and $\angle B=\angle B$ (common) $\Rightarrow$ AA.
  • (iii) $\triangle AEP\sim\triangle ADB$: $\angle AEP=\angle ADB=90^\circ$ and $\angle A=\angle A$ (common) $\Rightarrow$ AA.
  • (iv) $\triangle PDC\sim\triangle BEC$: $\angle PDC=\angle BEC=90^\circ$ and $\angle C=\angle C$ (common) $\Rightarrow$ AA.

Q8. $E$ on side $AD$ produced of parallelogram $ABCD$; $BE$ meets $CD$ at $F$. Show $\triangle ABE\sim\triangle CFB$. In a parallelogram $\angle A=\angle C$ (opposite angles). Also $AE\parallel BC\Rightarrow\angle AEB=\angle CBF$ (alternate angles). By AA, $\triangle ABE\sim\triangle CFB$.

Q9. $\triangle ABC$ (right at $B$) and $\triangle AMP$ (right at $M$). (i) $\angle ABC=\angle AMP=90^\circ$ and $\angle A=\angle A$ (common) $\Rightarrow\triangle ABC\sim\triangle AMP$ (AA). (ii) From similarity $\dfrac{CA}{PA}=\dfrac{BC}{MP}$.

Q10. $CD,GH$ bisect $\angle ACB,\angle EGF$; $\triangle ABC\sim\triangle FEG$. (i) Since the triangles are similar, $\angle C=\angle G$, so the halves $\angle ACD=\angle FGH$; with $\angle A=\angle F$, $\triangle ACD\sim\triangle FGH$ (AA), giving $\dfrac{CD}{GH}=\dfrac{AC}{FG}$. (ii) $\triangle DCB\sim\triangle HGE$: $\angle DCB=\angle HGE$ (halves) and $\angle B=\angle E\Rightarrow$ AA. (iii) $\triangle DCA\sim\triangle HGF$: $\angle DCA=\angle HGF$ (halves) and $\angle A=\angle F\Rightarrow$ AA.

Q11. Isosceles $\triangle ABC$ with $AB=AC$, $E$ on $CB$ produced, $AD\perp BC$, $EF\perp AC$; prove $\triangle ABD\sim\triangle ECF$. Since $AB=AC$, $\angle ABD=\angle ACB=\angle ECF$. Also $\angle ADB=\angle EFC=90^\circ$. By AA, $\triangle ABD\sim\triangle ECF$.

Q12. Sides $AB,BC$ and median $AD$ of $\triangle ABC$ are proportional to sides $PQ,QR$ and median $PM$ of $\triangle PQR$; show $\triangle ABC\sim\triangle PQR$. $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$. Since $BD=\tfrac12BC$ and $QM=\tfrac12QR$, $\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$, so $\triangle ABD\sim\triangle PQM$ (SSS) $\Rightarrow\angle B=\angle Q$. With $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and $\angle B=\angle Q$, SAS gives $\triangle ABC\sim\triangle PQR$.

Q13. $D$ on $BC$ with $\angle ADC=\angle BAC$; show $CA^2=CB\cdot CD$. In $\triangle ADC$ and $\triangle BAC$: $\angle ADC=\angle BAC$ (given) and $\angle C=\angle C$ (common). By AA, $\triangle ADC\sim\triangle BAC$, so $\dfrac{CA}{CB}=\dfrac{CD}{CA}\Rightarrow CA^2=CB\cdot CD$.

Q14. Sides $AB,AC$ and median $AD$ of $\triangle ABC$ proportional to sides $PQ,PR$ and median $PM$ of $\triangle PQR$; show $\triangle ABC\sim\triangle PQR$. Produce $AD$ to $E$ with $AD=DE$ and $PM$ to $N$ with $PM=MN$; the resulting parallelograms give $\triangle ABE'$-type relations that reduce (via SSS on the doubled triangles) to $\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}$, hence $\triangle ABC\sim\triangle PQR$ (SSS).

Q15. Pole and tower shadows. A $6$ m pole casts a $4$ m shadow when a tower casts a $28$ m shadow. The Sun's rays make equal angles, so the triangles are similar: $\dfrac{\text{height of tower}}{\text{height of pole}}=\dfrac{\text{tower shadow}}{\text{pole shadow}}\Rightarrow h=\dfrac{6\times28}{4}=42$ m.

Q16. $AD,PM$ medians of $\triangle ABC\sim\triangle PQR$; prove $\dfrac{AB}{PQ}=\dfrac{AD}{PM}$. Similarity gives $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ and $\angle B=\angle Q$. As $BD=\tfrac12BC,\ QM=\tfrac12QR$, $\dfrac{AB}{PQ}=\dfrac{BD}{QM}$. With $\angle B=\angle Q$, SAS gives $\triangle ABD\sim\triangle PQM$, hence $\dfrac{AB}{PQ}=\dfrac{AD}{PM}$.

13. Common mistakes to avoid

  • Writing the similarity with the wrong vertex order — $\triangle ABC\sim\triangle DEF$ is not the same as $\triangle ABC\sim\triangle EDF$.
  • Calling two figures similar from one condition only — for polygons you need equal angles and proportional sides together.
  • In SAS similarity, using sides that do not include the equal angle — the angle must be between the two proportional sides.
  • Confusing BPT with its converse: BPT starts from "parallel" to get a ratio; the converse starts from a ratio to prove "parallel".
  • Forgetting to justify each step (alternate angles, vertically opposite, common angle) in a proof — board markers award method marks.
  • Mixing up correspondence when reading off ratios from $\triangle ABC\sim\triangle PQR$: it is $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}$.

14. Quick revision checklist

  • Similar = same shape; congruent $\Rightarrow$ similar, but not conversely.
  • Polygons similar $\iff$ equal corresponding angles AND proportional corresponding sides.
  • BPT: $DE\parallel BC\Rightarrow\dfrac{AD}{DB}=\dfrac{AE}{EC}$; converse proves a line parallel.
  • Similarity criteria: AAA/AA, SSS, SAS (angle must be included), plus RHS for right triangles.
  • From $\triangle ABC\sim\triangle PQR$: ratio of any matching segments (sides, medians, altitudes, bisectors) equals the scale factor.
  • Indirect measurement (heights, shadows, distances) uses AA-similar triangles.
Practice MCQs
1. Which statement is always true?
  1. All similar figures are congruent
  2. All congruent figures are similar
  3. All rectangles are similar
  4. All isosceles triangles are similar
Answer: (B) congruent figures have the same shape and size, so they are certainly similar.
2. In $\triangle ABC$, $DE\parallel BC$ with $AD=2,\ DB=3,\ AE=4$. Then $EC=$
  1. $6$
  2. $5$
  3. $8$
  4. $\dfrac{8}{3}$
Answer: (A) $\dfrac{AD}{DB}=\dfrac{AE}{EC}\Rightarrow\dfrac23=\dfrac{4}{EC}\Rightarrow EC=6.$
3. If $\triangle ABC\sim\triangle DEF$ and $\dfrac{AB}{DE}=\dfrac{2}{3}$, then $\dfrac{BC}{EF}=$
  1. $\dfrac32$
  2. $\dfrac23$
  3. $\dfrac49$
  4. cannot say
Answer: (B) all corresponding sides share the same ratio, so $\dfrac{BC}{EF}=\dfrac23.$
4. Two triangles are similar if their corresponding:
  1. angles are equal only
  2. sides are equal
  3. sides are proportional
  4. areas are equal
Answer: (C) proportional sides (SSS) imply similarity; for angles, two equal angles (AA) suffice.
5. In $\triangle ABC$, a line divides $AB,AC$ with $\dfrac{AD}{DB}=\dfrac{AE}{EC}$. Then $DE$ is:
  1. perpendicular to $BC$
  2. parallel to $BC$
  3. equal to $BC$
  4. a median
Answer: (B) by the converse of BPT, $DE\parallel BC$.
6. Which criterion needs the equal angle to lie between the proportional sides?
  1. AAA
  2. SSS
  3. SAS
  4. AA
Answer: (C) in SAS the angle must be included between the two proportional sides.
7. A vertical pole $6$ m high casts a shadow $4$ m; a tower's shadow is $28$ m. Tower height $=$
  1. $36$ m
  2. $42$ m
  3. $48$ m
  4. $18$ m
Answer: (B) similar triangles: $h=\dfrac{6\times28}{4}=42$ m.
8. If $D$ is on $BC$ of $\triangle ABC$ with $\angle ADC=\angle BAC$, then:
  1. $AB^2=BC\cdot CD$
  2. $CA^2=CB\cdot CD$
  3. $AD^2=BD\cdot DC$
  4. $CA^2=AB\cdot BC$
Answer: (B) $\triangle ADC\sim\triangle BAC$ gives $CA^2=CB\cdot CD$.
9. In $\triangle POQ\sim\triangle SOR$ formed by $PQ\parallel RS$, the criterion used is:
  1. SSS
  2. SAS
  3. AA
  4. RHS
Answer: (C) alternate angles equal plus vertically opposite angles equal give AA similarity.
10. If two triangles are similar with scale factor $\dfrac{3}{5}$, then the ratio of their corresponding medians is:
  1. $\dfrac{9}{25}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{5}{3}$
  4. $1$
Answer: (B) corresponding medians (and all matching segments) are in the same ratio as the sides, $\dfrac35$.
Assertion–Reason
A: A line parallel to one side of a triangle divides the other two sides in the same ratio.   R: This is the Basic Proportionality (Thales) Theorem.
Answer: Both A and R are true, and R correctly names/explains A — A is exactly the statement of Theorem 6.1.
A: All squares are similar.   R: All rectangles are similar.
Answer: A is true (squares have equal angles and proportional sides), but R is false — rectangles can have different length-to-breadth ratios, so they need not be similar.
Previous-year questions
Q1. State and prove the Basic Proportionality Theorem. (CBSE, 3-4 marks)
Outline: In $\triangle ABC$ with $DE\parallel BC$, join $BE,CD$ and drop $DM\perp AC,\ EN\perp AB$. Then $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)}=\dfrac{AD}{DB}$ and $\dfrac{\text{ar}(ADE)}{\text{ar}(DEC)}=\dfrac{AE}{EC}$; since $\text{ar}(BDE)=\text{ar}(DEC)$ (same base $DE$, same parallels), $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.
Q2. In $\triangle ABC$, $DE\parallel BC$. If $AD=x,\ DB=x-2,\ AE=x+2,\ EC=x-1$, find $x$. (CBSE, 3 marks)
Answer: By BPT $\dfrac{x}{x-2}=\dfrac{x+2}{x-1}\Rightarrow x(x-1)=(x-2)(x+2)\Rightarrow x^2-x=x^2-4\Rightarrow x=4.$
Q3. $D$ is a point on $BC$ of $\triangle ABC$ such that $\angle ADC=\angle BAC$. Prove $CA^2=CB\cdot CD$. (CBSE, 3 marks)
Outline: In $\triangle ADC$ and $\triangle BAC$, $\angle ADC=\angle BAC$ and $\angle C$ common $\Rightarrow\triangle ADC\sim\triangle BAC$ (AA), so $\dfrac{CA}{CB}=\dfrac{CD}{CA}\Rightarrow CA^2=CB\cdot CD.$
Q4. Diagonals of a trapezium $ABCD$ with $AB\parallel DC$ meet at $O$. Prove $\dfrac{AO}{OC}=\dfrac{BO}{OD}$. (CBSE, 3 marks)
Outline: In $\triangle OAB$ and $\triangle OCD$: alternate angles $\angle OAB=\angle OCD,\ \angle OBA=\angle ODC$ and vertically opposite $\angle AOB=\angle COD$, so $\triangle OAB\sim\triangle OCD$ (AAA) $\Rightarrow\dfrac{AO}{OC}=\dfrac{BO}{OD}.$
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More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Arithmetic Progressions · Coordinate Geometry · Introduction to Trigonometry · Some Applications of Trigonometry