Triangles — CBSE Class 10 Maths Textbook Solutions

(i) similar — circles can have different radii, so they are not all congruent, but they always have the same shape.

(ii) similar — for the same reason as circles; two squares with different side lengths are similar but not congruent.

(iii) equilateral — every equilateral triangle has all angles equal to $60^\circ$60^ and sides in the same ratio, so equilateral triangles are always similar (unlike general isosceles triangles).

(iv) (a) equal, (b) proportional.

(i) Similar figures:

1. Any two squares of different side lengths — same shape (all angles $90^\circ$90^), sides in the same ratio.
2. Any two equilateral triangles of different side lengths — same shape (all angles $60^\circ$60^).
   (Other valid answers: two circles of different radii; a photograph and its enlarged print.)

(ii) Non-similar figures:

1. A triangle and a square — different shape entirely (different number of sides).
2. A rectangle that is not a square, and a rhombus that is not a square — corresponding angles are not all equal even where some side ratios might match.
   (Other valid answer: a scalene triangle and an equilateral triangle.)

Compare the two conditions needed for similarity of polygons.

Corresponding sides: $\frac{PQ}{AB}=\frac{QR}{BC}=\frac{RS}{CD}=\frac{SP}{DA}=\frac{3}{6}=\frac{1}{2}$PQ/AB=QR/BC=RS/CD=SP/DA=3/6=1/2 — all four ratios are equal, so the sides are proportional.

Corresponding angles: the square has all angles $90^\circ$90^, while the rhombus has angles $70^\circ$70^ and $110^\circ$110^. The corresponding angles are not equal.

Since both conditions must hold for polygons to be similar, and the angle condition fails here, $PQRS$PQRS and $ABCD$ABCD are not similar — No.

By the Basic Proportionality Theorem, since $DE\parallel BC$DE BC in $\triangle ABC$ABC: $\frac{AD}{DB}=\frac{AE}{EC}$AD/DB=AE/EC.

(i) $AD=1.5\text{ cm}$AD=1.5 cm, $DB=3\text{ cm}$DB=3 cm, $AE=1\text{ cm}$AE=1 cm.
$\frac{1.5}{3}=\frac{1}{EC}\Rightarrow EC=\frac{1\times3}{1.5}=2\text{ cm}$1.5/3=1/EC EC=1×3/1.5=2 cm.

(ii) $DB=7.2\text{ cm}$DB=7.2 cm, $AE=1.8\text{ cm}$AE=1.8 cm, $EC=5.4\text{ cm}$EC=5.4 cm.
$\frac{AD}{7.2}=\frac{1.8}{5.4}=\frac{1}{3}\Rightarrow AD=\frac{7.2}{3}=2.4\text{ cm}$AD/7.2=1.8/5.4=1/3 AD=7.2/3=2.4 cm.

By the converse of the Basic Proportionality Theorem, $EF\parallel QR$EF QR if and only if $\frac{PE}{EQ}=\frac{PF}{FR}$PE/EQ=PF/FR.

(i) $\frac{PE}{EQ}=\frac{3.9}{3}=1.3$PE/EQ=3.9/3=1.3; $\frac{PF}{FR}=\frac{3.6}{2.4}=1.5$PF/FR=3.6/2.4=1.5. These are not equal, so $EF$EF is not parallel to $QR$QR.

(ii) $\frac{PE}{QE}=\frac{4}{4.5}=\frac{8}{9}$PE/QE=4/4.5=8/9; $\frac{PF}{RF}=\frac{8}{9}$PF/RF=8/9. These are equal, so $EF\parallel QR$EF QR — Yes.

(iii) $EQ=PQ-PE=1.28-0.18=1.10\text{ cm}$EQ=PQ-PE=1.28-0.18=1.10 cm; $FR=PR-PF=2.56-0.36=2.20\text{ cm}$FR=PR-PF=2.56-0.36=2.20 cm.
$\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{9}{55}$PE/EQ=0.18/1.10=9/55; $\frac{PF}{FR}=\frac{0.36}{2.20}=\frac{9}{55}$PF/FR=0.36/2.20=9/55. These are equal, so $EF\parallel QR$EF QR — Yes.

In $\triangle ABC$ABC, since $LM\parallel CB$LM CB, by the Basic Proportionality Theorem (Theorem 6.1):
$\frac{AM}{AB}=\frac{AL}{AC}\qquad\ldots(1)$AM/AB=AL/AC(1)

In $\triangle ACD$ACD, since $LN\parallel CD$LN CD, by the Basic Proportionality Theorem:
$\frac{AN}{AD}=\frac{AL}{AC}\qquad\ldots(2)$AN/AD=AL/AC(2)

From (1) and (2), both ratios equal $\frac{AL}{AC}$AL/AC, so:
$\frac{AM}{AB}=\frac{AN}{AD}.$AM/AB=AN/AD.
Hence proved.

In $\triangle ABE$ABE, since $DF\parallel AE$DF AE, by Theorem 6.1:
$\frac{BF}{FE}=\frac{BD}{DA}\qquad\ldots(1)$BF/FE=BD/DA(1)

In $\triangle ABC$ABC, since $DE\parallel AC$DE AC, by Theorem 6.1:
$\frac{BD}{DA}=\frac{BE}{EC}\qquad\ldots(2)$BD/DA=BE/EC(2)

From (1) and (2):
$\frac{BF}{FE}=\frac{BD}{DA}=\frac{BE}{EC}.$BF/FE=BD/DA=BE/EC.
Hence, $\frac{BF}{FE}=\frac{BE}{EC}$BF/FE=BE/EC.

In $\triangle POQ$POQ, since $DE\parallel OQ$DE OQ, by Theorem 6.1:
$\frac{PD}{DO}=\frac{PE}{EQ}\qquad\ldots(1)$PD/DO=PE/EQ(1)

In $\triangle POR$POR, since $DF\parallel OR$DF OR, by Theorem 6.1:
$\frac{PD}{DO}=\frac{PF}{FR}\qquad\ldots(2)$PD/DO=PF/FR(2)

From (1) and (2): $\frac{PE}{EQ}=\frac{PF}{FR}$PE/EQ=PF/FR.

In $\triangle PQR$PQR, since $E$E on $PQ$PQ and $F$F on $PR$PR divide the sides in the same ratio, by the converse of the Basic Proportionality Theorem (Theorem 6.2), $EF\parallel QR$EF QR.

In $\triangle OPQ$OPQ, since $AB\parallel PQ$AB PQ, by Theorem 6.1:
$\frac{OA}{AP}=\frac{OB}{BQ}\qquad\ldots(1)$OA/AP=OB/BQ(1)

In $\triangle OPR$OPR, since $AC\parallel PR$AC PR, by Theorem 6.1:
$\frac{OA}{AP}=\frac{OC}{CR}\qquad\ldots(2)$OA/AP=OC/CR(2)

From (1) and (2): $\frac{OB}{BQ}=\frac{OC}{CR}$OB/BQ=OC/CR.

In $\triangle OQR$OQR, since $B$B on $OQ$OQ and $C$C on $OR$OR divide the sides in the same ratio, by the converse of the Basic Proportionality Theorem (Theorem 6.2), $BC\parallel QR$BC QR.

Given: $\triangle ABC$ABC with $D$D the mid-point of $AB$AB, and $DE\parallel BC$DE BC meeting $AC$AC at $E$E.
To prove: $E$E is the mid-point of $AC$AC, i.e., $AE=EC$AE=EC.

Proof: Since $D$D is the mid-point of $AB$AB, $AD=DB$AD=DB, so $\frac{AD}{DB}=1$AD/DB=1.

Since $DE\parallel BC$DE BC in $\triangle ABC$ABC, by the Basic Proportionality Theorem (Theorem 6.1):
$\frac{AD}{DB}=\frac{AE}{EC}.$AD/DB=AE/EC.

Since $\frac{AD}{DB}=1$AD/DB=1, it follows that $\frac{AE}{EC}=1$AE/EC=1, i.e., $AE=EC$AE=EC.

Hence $E$E is the mid-point of $AC$AC — the line bisects the third side.

Given: $\triangle ABC$ABC with $D$D, $E$E the mid-points of $AB$AB, $AC$AC respectively.
To prove: $DE\parallel BC$DE BC.

Proof: Since $D$D is the mid-point of $AB$AB, $AD=DB\Rightarrow\frac{AD}{DB}=1$AD=DB/DB=1.
Since $E$E is the mid-point of $AC$AC, $AE=EC\Rightarrow\frac{AE}{EC}=1$AE=EC/EC=1.

So $\frac{AD}{DB}=\frac{AE}{EC}$AD/DB=AE/EC (each equal to 1).

By the converse of the Basic Proportionality Theorem (Theorem 6.2), since the line $DE$DE divides sides $AB$AB and $AC$AC in the same ratio, $DE\parallel BC$DE BC.

In $\triangle AOB$AOB and $\triangle COD$COD:

$\angle AOB=\angle COD$AOB= COD (vertically opposite angles)

$\angle OAB=\angle OCD$OAB= OCD (alternate angles, since $AB\parallel DC$AB DC and $AC$AC is a transversal)

By the AA similarity criterion, $\triangle AOB\sim\triangle COD$AOB COD.

So their corresponding sides are proportional:
$\frac{AO}{CO}=\frac{BO}{DO}.$AO/CO=BO/DO.

Cross-multiplying: $AO\cdot DO=CO\cdot BO$AO· DO=CO· BO. Dividing both sides by $BO\cdot DO$BO· DO:
$\frac{AO}{BO}=\frac{CO}{DO}.$AO/BO=CO/DO.
Hence proved.

Given: $\frac{AO}{BO}=\frac{CO}{DO}$AO/BO=CO/DO, which can be rewritten as $\frac{AO}{CO}=\frac{BO}{DO}$AO/CO=BO/DO.

In $\triangle AOB$AOB and $\triangle COD$COD:

$\angle AOB=\angle COD$AOB= COD (vertically opposite angles)

$\frac{AO}{CO}=\frac{BO}{DO}$AO/CO=BO/DO (given, rearranged) — the sides including the equal angles are proportional.

By the SAS similarity criterion, $\triangle AOB\sim\triangle COD$AOB COD.

So $\angle OAB=\angle OCD$OAB= OCD (corresponding angles of similar triangles).

But $\angle OAB$OAB and $\angle OCD$OCD are alternate angles formed by transversal $AC$AC with lines $AB$AB and $DC$DC. Since these alternate angles are equal, $AB\parallel DC$AB DC.

Hence $ABCD$ABCD is a trapezium (with $AB\parallel DC$AB DC).

(i) All three corresponding angles are equal ($\angle A=\angle P,\ \angle B=\angle Q,\ \angle C=\angle R$A= P, B= Q, C= R). Similar, AAA criterion: $\triangle ABC\sim\triangle PQR$ABC PQR.

(ii) Check the ratio of corresponding sides: $\frac{AB}{QR}=\frac{3}{6}=\frac{1}{2}$AB/QR=3/6=1/2, $\frac{BC}{RP}=\frac{4}{8}=\frac{1}{2}$BC/RP=4/8=1/2, $\frac{CA}{PQ}=\frac{5}{10}=\frac{1}{2}$CA/PQ=5/10=1/2. All equal. Similar, SSS criterion: $\triangle ABC\sim\triangle QRP$ABC QRP.

(iii) $\frac{AB}{PQ}=\frac{3}{6}=0.5$AB/PQ=3/6=0.5, $\frac{BC}{QR}=\frac{4}{9}\approx0.44$BC/QR=4/90.44, $\frac{CA}{RP}=\frac{6}{8}=0.75$CA/RP=6/8=0.75. These are not equal. Not similar.

(iv) $\angle M=\angle Q=100^\circ$M= Q=100^ (the included angle), and $\frac{MN}{QP}=\frac{2}{4}=0.5$MN/QP=2/4=0.5, $\frac{ML}{QR}=\frac{3.5}{7}=0.5$ML/QR=3.5/7=0.5 — equal ratios about the equal included angle. Similar, SAS criterion: $\triangle MNL\sim\triangle QPR$MNL QPR.

(v) $\angle A=\angle D=90^\circ$A= D=90^, but $\frac{AB}{DE}=\frac{5}{6}$AB/DE=5/6 while $\frac{AC}{DF}=\frac{12}{9}=\frac{4}{3}$AC/DF=12/9=4/3 — the sides about the equal angle are not in the same ratio. Not similar.

(vi) $\angle D=\angle P=45^\circ$D= P=45^ and $\angle E=\angle Q=55^\circ$E= Q=55^ — two pairs of corresponding angles equal. Similar, AA criterion: $\triangle DEF\sim\triangle PQR$DEF PQR.

Since $BD$BD is a straight line, $\angle BOC$BOC and $\angle DOC$DOC form a linear pair:
$\angle DOC=180^\circ-\angle BOC=180^\circ-125^\circ=55^\circ.$DOC=180^- BOC=180^-125^=55^.

In $\triangle ODC$ODC, the angles sum to $180^\circ$180^:
$\angle DCO=180^\circ-\angle CDO-\angle DOC=180^\circ-70^\circ-55^\circ=55^\circ.$DCO=180^- CDO- DOC=180^-70^-55^=55^.

Since $\triangle ODC\sim\triangle OBA$ODC OBA, corresponding angles are equal; $\angle DCO$DCO corresponds to $\angle OAB$OAB:
$\angle OAB=\angle DCO=55^\circ.$OAB= DCO=55^.

So $\angle DOC=55^\circ$DOC=55^, $\angle DCO=55^\circ$DCO=55^, $\angle OAB=55^\circ$OAB=55^.

In $\triangle OAB$OAB and $\triangle OCD$OCD:

$\angle AOB=\angle COD$AOB= COD (vertically opposite angles)

$\angle OAB=\angle OCD$OAB= OCD (alternate angles, $AB\parallel DC$AB DC, transversal $AC$AC)

By the AA similarity criterion, $\triangle OAB\sim\triangle OCD$OAB OCD.

Since corresponding sides of similar triangles are proportional:
$\frac{OA}{OC}=\frac{OB}{OD}.$OA/OC=OB/OD.
Hence proved.

Step 1. Since $\angle PQR=\angle PRQ$PQR= PRQ (given, $\angle1=\angle2$1=2), the sides opposite these equal angles are equal:
$PR=PQ\qquad\ldots(1)$PR=PQ(1)

Step 2. Substitute (1) into the given relation:
$\frac{QR}{QS}=\frac{QT}{PR}=\frac{QT}{PQ}$QR/QS=QT/PR=QT/PQ

Rearranging (cross-multiplying and regrouping):
$\frac{PQ}{QT}=\frac{QS}{QR}\qquad\ldots(2)$PQ/QT=QS/QR(2)

Step 3. In $\triangle PQS$PQS and $\triangle TQR$TQR:

$\angle PQS=\angle TQR$PQS= TQR (the same angle $\angle Q$Q, since $T$T lies on $QP$QP and $S$S lies on $QR$QR)

$\frac{PQ}{TQ}=\frac{QS}{QR}$PQ/TQ=QS/QR [from (2)] — sides including the equal angle are proportional

By the SAS similarity criterion, $\triangle PQS\sim\triangle TQR$PQS TQR. Hence proved.

In $\triangle RPQ$RPQ and $\triangle RTS$RTS:

$\angle RPQ=\angle RTS$RPQ= RTS (given, $\angle P=\angle RTS$P= RTS)

$\angle PRQ=\angle TRS$PRQ= TRS (the same angle $\angle R$R, common to both triangles)

By the AA similarity criterion, $\triangle RPQ\sim\triangle RTS$RPQ RTS. Hence proved.

Since $\triangle ABE\cong\triangle ACD$ABE ACD, corresponding sides are equal:
$AB=AC\qquad\text{and}\qquad AE=AD.$AB=AC AE=AD.

In $\triangle ADE$ADE and $\triangle ABC$ABC:

$\frac{AD}{AB}=\frac{AE}{AC}$AD/AB=AE/AC — since $AD=AE$AD=AE and $AB=AC$AB=AC, both ratios equal $\frac{AD}{AB}$AD/AB

$\angle DAE=\angle BAC$DAE= BAC (the same angle at vertex $A$A)

By the SAS similarity criterion, $\triangle ADE\sim\triangle ABC$ADE ABC. Hence proved.

(i) $\triangle AEP\sim\triangle CDP$AEP CDP:
$\angle AEP=\angle CDP=90^\circ$AEP= CDP=90^ ($AD\perp BC$AD BC and $CE\perp AB$CE AB are altitudes)
$\angle APE=\angle CPD$APE= CPD (vertically opposite angles)
By AA similarity, $\triangle AEP\sim\triangle CDP$AEP CDP.

(ii) $\triangle ABD\sim\triangle CBE$ABD CBE:
$\angle ADB=\angle CEB=90^\circ$ADB= CEB=90^ (altitudes)
$\angle ABD=\angle CBE$ABD= CBE (the same angle $\angle B$B, common to both triangles)
By AA similarity, $\triangle ABD\sim\triangle CBE$ABD CBE.

(iii) $\triangle AEP\sim\triangle ADB$AEP ADB:
$\angle AEP=\angle ADB=90^\circ$AEP= ADB=90^ (altitudes)
$\angle PAE=\angle BAD$PAE= BAD (the same angle $\angle A$A, common to both triangles, since $P$P lies on $AD$AD and $E$E lies on $AB$AB)
By AA similarity, $\triangle AEP\sim\triangle ADB$AEP ADB.

(iv) $\triangle PDC\sim\triangle BEC$PDC BEC:
$\angle PDC=\angle BEC=90^\circ$PDC= BEC=90^ (altitudes)
$\angle PCD=\angle BCE$PCD= BCE (the same angle $\angle C$C, common to both triangles, since $P$P lies on $CE$CE and $D$D lies on $BC$BC)
By AA similarity, $\triangle PDC\sim\triangle BEC$PDC BEC.

Since $ABCD$ABCD is a parallelogram, $AB\parallel DC$AB DC and $AD\parallel BC$AD BC, and opposite angles are equal: $\angle A=\angle C$A= C.

Since $E$E lies on $AD$AD produced, $\angle EAB=\angle DAB=\angle A$EAB= DAB= A; since $F$F lies on $DC$DC, $\angle BCF=\angle BCD=\angle C$BCF= BCD= C. As $\angle A=\angle C$A= C:
$\angle EAB=\angle BCF\qquad\ldots(1)$EAB= BCF(1)

Since $AD\parallel BC$AD BC (opposite sides of the parallelogram) and line $BFE$BFE is a transversal cutting both:
$\angle AEB=\angle EBC\ \text{(alternate angles)}$AEB= EBC (alternate angles)
and since $F$F lies on segment $BE$BE, $\angle EBC=\angle FBC$EBC= FBC, so
$\angle AEB=\angle CBF\qquad\ldots(2)$AEB= CBF(2)

From (1) and (2), two pairs of angles of $\triangle ABE$ABE and $\triangle CFB$CFB are equal, so by the AA similarity criterion:
$\triangle ABE\sim\triangle CFB.$ABE CFB.
Hence proved.

In $\triangle ABC$ABC and $\triangle AMP$AMP:

$\angle ABC=\angle AMP=90^\circ$ABC= AMP=90^ (given: right angles at $B$B and $M$M)

$\angle BAC=\angle MAP$BAC= MAP (the same angle at vertex $A$A, common to both triangles)

(i) By the AA similarity criterion, $\triangle ABC\sim\triangle AMP$ABC AMP.

(ii) Since the triangles are similar (correspondence $A\leftrightarrow A,\ B\leftrightarrow M,\ C\leftrightarrow P$A A, B M, C P), their corresponding sides are proportional:
$\frac{AB}{AM}=\frac{BC}{MP}=\frac{CA}{PA}.$AB/AM=BC/MP=CA/PA.
In particular, $\frac{CA}{PA}=\frac{BC}{MP}$CA/PA=BC/MP. Hence proved.

Since $\triangle ABC\sim\triangle FEG$ABC FEG: $\angle A=\angle F$A= F, $\angle B=\angle E$B= E, $\angle ACB=\angle FGE$ACB= FGE (call this common angle $\angle C=\angle G$C= G).

Since $CD$CD bisects $\angle ACB$ACB: $\angle ACD=\angle DCB=\frac{\angle C}{2}$ACD= DCB= C/2.
Since $GH$GH bisects $\angle EGF$EGF: $\angle FGH=\angle HGE=\frac{\angle G}{2}=\frac{\angle C}{2}$FGH= HGE= G/2= C/2 (as $\angle C=\angle G$C= G).

So $\angle ACD=\angle FGH\qquad\ldots(*)$ACD= FGH(*) and $\angle DCB=\angle HGE\qquad\ldots(**)$DCB= HGE(**)** (iii)** In $\triangle DCA$DCA and $\triangle HGF$HGF: $\angle A=\angle F$A= F (given) and $\angle ACD=\angle FGH$ACD= FGH [from $(*)$(*)]. By AA similarity, $\triangle DCA\sim\triangle HGF$DCA HGF.

(i) Since $\triangle DCA\sim\triangle HGF$DCA HGF (part iii), corresponding sides are proportional:
$\frac{CD}{GH}=\frac{AC}{FG}\ \left(=\frac{DA}{HF}\right).$CD/GH=AC/FG (=DA/HF).

(ii) In $\triangle DCB$DCB and $\triangle HGE$HGE: $\angle B=\angle E$B= E (given) and $\angle DCB=\angle HGE$DCB= HGE [from $(**)$(**)]. By AA similarity, $\triangle DCB\sim\triangle HGE$DCB HGE.

Since $AB=AC$AB=AC in $\triangle ABC$ABC, the base angles are equal: $\angle ABC=\angle ACB$ABC= ACB (angles opposite equal sides).

Since $D$D lies on $BC$BC, $\angle ABD=\angle ABC$ABD= ABC.

Since $E$E lies on $CB$CB produced (beyond $B$B), ray $CE$CE is the same as ray $CB$CB, and $F$F lies on $CA$CA, so $\angle ECF=\angle BCA=\angle ACB$ECF= BCA= ACB (same angle at $C$C).

So $\angle ABD=\angle ABC=\angle ACB=\angle ECF$ABD= ABC= ACB= ECF, i.e.
$\angle ABD=\angle ECF\qquad\ldots(1)$ABD= ECF(1)

Also, $\angle ADB=\angle EFC=90^\circ$ADB= EFC=90^ (given $AD\perp BC$AD BC and $EF\perp AC$EF AC) $\qquad\ldots(2)$(2)

From (1) and (2), by the AA similarity criterion:
$\triangle ABD\sim\triangle ECF.$ABD ECF.
Hence proved.

Given: $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$AB/PQ=BC/QR=AD/PM, where $AD$AD, $PM$PM are medians.

Since $AD$AD is a median, $D$D is the mid-point of $BC$BC, so $BD=\frac{BC}{2}$BD=BC/2. Similarly $QM=\frac{QR}{2}$QM=QR/2. So:
$\frac{BD}{QM}=\frac{BC}{QR}=\frac{AB}{PQ}=\frac{AD}{PM}$BD/QM=BC/QR=AB/PQ=AD/PM

In $\triangle ABD$ABD and $\triangle PQM$PQM: all three corresponding sides are in the same ratio ($\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$AB/PQ=BD/QM=AD/PM). By the SSS similarity criterion, $\triangle ABD\sim\triangle PQM$ABD PQM.

So $\angle ABD=\angle PQM$ABD= PQM, i.e. $\angle ABC=\angle PQR$ABC= PQR (since $D$D lies on $BC$BC and $M$M on $QR$QR).