Similar Triangles PYQs
Overview
This page provides comprehensive Class 10 Maths Similar Triangles PYQs | Triangles. Similar Triangles previous year questions for Class 10 Maths Triangles. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Concept-wise CBSE Class 10 Questions with Step-by-Step Solutions
Qsim1
2020
00:00
All concentric circles are _______ to each other.
Concentric circles have the same center and identical shape, differing only in radius.
Therefore, all concentric circles are similar to each other.
Final Answer: similar
Qsim2
2026
00:00
It is given that ΔABC ~ ΔQRP such that AB = 9 cm, BC = 5 cm and PR = 2 cm. Length of side QR is:
Step 1: Since $\triangle ABC \sim \triangle QRP$, corresponding sides are in ratio:
$$\frac{AB}{QR} = \frac{BC}{RP}$$
Step 2: Substitute $AB = 9$, $BC = 5$, $PR = 2$:
$$\frac{9}{QR} = \frac{5}{2} \implies 5 QR = 18 \implies QR = 3.6\text{ cm}$$
Final Answer: 3.6 cm (Option d)
Qsim3
2026
00:00
If ΔABC and ΔDEF are similar such that 2AB = DE and BC = 8 cm, then EF is equal to:
Step 1: $\triangle ABC \sim \triangle DEF \implies \frac{AB}{DE} = \frac{BC}{EF}$.
Step 2: Since $DE = 2 AB$, $\frac{1}{2} = \frac{8}{EF} \implies EF = 16\text{ cm}$.
Final Answer: 16 cm (Option d)
Qsim4
2025
00:00
Given ΔABC ~ ΔPQR, ∠A = 30° and ∠Q = 90°. The value of (∠R + ∠B) is:
Step 1: $\triangle ABC \sim \triangle PQR \implies \angle A = \angle P = 30^\circ, \angle B = \angle Q = 90^\circ, \angle C = \angle R$.
Step 2: In $\triangle PQR$, $\angle R = 180^\circ - (30^\circ + 90^\circ) = 60^\circ$.
Step 3: $\angle R + \angle B = 60^\circ + 90^\circ = 150^\circ$.
Final Answer: 150° (Option c)
Qsim5
2023
00:00
If ΔABC ~ ΔPQR with ∠A = 32° and ∠R = 65°, then the measure of ∠B is:
Step 1: Corresponding angles are equal: $\angle C = \angle R = 65^\circ$ and $\angle A = 32^\circ$.
Step 2: $\angle B = 180^\circ - (32^\circ + 65^\circ) = 83^\circ$.
Final Answer: 83° (Option c)
Qsim6
2023
00:00
In the given figure, DE ∥ BC. The value of x is:
Step 1: $\triangle ADE \sim \triangle ABC \implies \frac{AD}{AB} = \frac{DE}{BC}$.
Step 2: Given $AD=2, DB=3 \implies AB=5$. Also $DE=4, BC=x$.
Step 3: $\frac{2}{5} = \frac{4}{x} \implies x = 10$.
Final Answer: 10 (Option d)
Qsim7
Term I, 2021-22
00:00
If ΔABC and ΔPQR are similar triangles such that ∠A = 31° and ∠R = 69°, then ∠Q is:
Step 1: $\angle P = \angle A = 31^\circ$ and $\angle R = 69^\circ$.
Step 2: In $\triangle PQR$, $\angle Q = 180^\circ - (31^\circ + 69^\circ) = 80^\circ$.
Final Answer: 80° (Option d)
Qsim8
2026
00:00
In the given figure, ΔAHK ~ ΔABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find the length of AC.
Step 1: Since $\triangle AHK \sim \triangle ABC$, corresponding sides are proportional:
$$\frac{AK}{AC} = \frac{HK}{BC}$$
Step 2: Substitute $AK = 10$, $BC = 3.5$, $HK = 7$:
$$\frac{10}{AC} = \frac{7}{3.5} = 2 \implies AC = 5\text{ cm}$$
Final Answer: 5 cm
Qsim9
Board Term I, 2017
00:00
In the figure, P is any point on side BC of ΔABC. PQ ∥ BA and PR ∥ CA are drawn. RQ is extended to meet BC produced at S. Prove that SP² = SB × SC.
Step 1: In $\triangle SBR$, $PQ \parallel BR \implies \frac{SP}{SB} = \frac{SQ}{SR}$.
Step 2: In $\triangle SQC$, $PR \parallel QC \implies \frac{SC}{SP} = \frac{SR}{SQ} \implies \frac{SP}{SC} = \frac{SQ}{SR}$.
Step 3: Equating both: $\frac{SP}{SB} = \frac{SC}{SP} \implies SP^2 = SB \times SC$. (Proved)
Final Answer: SP² = SB × SC (Proved)
Qsim10
2026
00:00
In the given figure, PQ ∥ YZ such that XP : PY = 2 : 3. If PQ = 5 cm, then YZ equals:
Step 1: $\triangle XPQ \sim \triangle XYZ \implies \frac{XP}{XY} = \frac{PQ}{YZ}$.
Step 2: $XY = XP + PY = 2 + 3 = 5$ parts.
Step 3: $\frac{2}{5} = \frac{5}{YZ} \implies 2 YZ = 25 \implies YZ = 12.5\text{ cm}$.
Final Answer: 12.5 cm (Option a)
Qsim11
2026
00:00
Devansh proved that ΔABC ~ ΔPQR using SAS similarity criteria. If he found ∠C = ∠R, then which of the following was proved true?
Step 1: For SAS similarity with included angle $\angle C = \angle R$, the sides containing these angles must be proportional.
Step 2: The sides containing $\angle C$ are $AC$ and $BC$, and sides containing $\angle R$ are $PR$ and $QR$.
Step 3: Thus $\frac{AC}{PR} = \frac{BC}{QR} \implies \frac{AC}{BC} = \frac{PR}{QR}$.
Final Answer: AC/BC = PR/QR (Option d)
Qsim12
2025
00:00
If in two triangles ΔDEF and ΔPQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
Step 1: $\triangle DEF \sim \triangle QRP$ by AA similarity.
Step 2: Corresponding ratio is $\frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP}$.
Step 3: Statement (d) $\frac{DE}{PQ} = \frac{EF}{RP}$ is NOT true.
Final Answer: Option (d)
Qsim13
2025
00:00
The measurements of ΔLMN and ΔABC are shown in the figure given below. The length of side AC is:
Step 1: In $\triangle LMN$, $\angle L = 180^\circ - (130^\circ + 28^\circ) = 22^\circ$.
Step 2: In $\triangle ABC$, $\angle C = 180^\circ - (130^\circ + 22^\circ) = 28^\circ$.
Step 3: Thus $\triangle LMN \sim \triangle ABC$ by AA similarity ($L \leftrightarrow A, M \leftrightarrow B, N \leftrightarrow C$).
Step 4: Ratio of sides: $\frac{LM}{AB} = \frac{45}{5} = 9 \implies \frac{LN}{AC} = 9 \implies \frac{72}{AC} = 9 \implies AC = 8\text{ cm}$.
Final Answer: 8 cm (Option c)
Qsim14
Board Term I, 2017
00:00
In the figure, if ΔBEA ≅ ΔCDA, then prove that ΔDEA ~ ΔBCA.
Step 1: Given $\triangle BEA \cong \triangle CDA \implies AE = AD$ and $AB = AC$ (CPCT).
Step 2: Therefore $\frac{AD}{AB} = \frac{AE}{AC}$.
Step 3: In $\triangle DEA$ and $\triangle BCA$, $\frac{AD}{AB} = \frac{AE}{AC}$ and $\angle A$ is common.
Step 4: Hence $\triangle DEA \sim \triangle BCA$ by SAS similarity criterion. (Proved)
Final Answer: ΔDEA ~ ΔBCA (Proved)
Qsim15
Board Term I, 2017
00:00
In ΔABC, ∠ADE = ∠B then prove that ΔADE ~ ΔABC. Also if AD = 7.6 cm, BD = 4.2 cm and BC = 8.4 cm, then find DE.
Step 1: In $\triangle ADE$ and $\triangle ABC$: $\angle A$ is common and $\angle ADE = \angle B$.
Step 2: By AA similarity, $\triangle ADE \sim \triangle ABC$.
Step 3: Ratio of sides: $\frac{AD}{AB} = \frac{DE}{BC}$.
Step 4: Here $AB = AD + BD = 7.6 + 4.2 = 11.8\text{ cm}$:
$$\frac{7.6}{11.8} = \frac{DE}{8.4} \implies DE = \frac{7.6 \times 8.4}{11.8} = 5.41\text{ cm (approx)}$$
Final Answer: DE ≈ 5.41 cm
Qsim16
2026
00:00
In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ~ ΔPQR, then prove that:
(i) ΔAMC ~ ΔPNR
(ii) ΔCMB ~ ΔRNQ
(i) ΔAMC ~ ΔPNR
(ii) ΔCMB ~ ΔRNQ
Step 1: $\triangle ABC \sim \triangle PQR \implies \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle A = \angle P, \angle B = \angle Q$.
Step 2: Since $M$ and $N$ are midpoints, $AB = 2 AM$ and $PQ = 2 PN$.
Step 3: So $\frac{2 AM}{2 PN} = \frac{AC}{PR} \implies \frac{AM}{PN} = \frac{AC}{PR}$.
Step 4: In $\triangle AMC$ and $\triangle PNR$, $\frac{AM}{PN} = \frac{AC}{PR}$ and $\angle A = \angle P \implies \triangle AMC \sim \triangle PNR$ (by SAS).
Step 5: Similarly, using $MB$ and $NQ$, $\triangle CMB \sim \triangle RNQ$. (Proved)
Final Answer: (i) ΔAMC ~ ΔPNR and (ii) ΔCMB ~ ΔRNQ (Proved)
Qsim17
2026
00:00
D is the mid-point of side BC of ΔABC. CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that
(i) OBGC is a parallelogram.
(ii) EF ∥ BC
(iii) ΔAEF ~ ΔABC
(i) OBGC is a parallelogram.
(ii) EF ∥ BC
(iii) ΔAEF ~ ΔABC
Step 1: In quadrilateral $OBGC$, diagonals $BC$ and $OG$ bisect each other at $D$ (since $BD=DC$ and $OD=DG$). Hence $OBGC$ is a parallelogram.
Step 2: Since $OBGC$ is a parallelogram, $BG \parallel OC \implies BG \parallel CE$ and $CG \parallel OB \implies CG \parallel BF$.
Step 3: In $\triangle ABG$, $OF \parallel BG \implies \frac{AF}{FB} = \frac{AO}{OG}$.
Step 4: In $\triangle ACG$, $OE \parallel CG \implies \frac{AE}{EC} = \frac{AO}{OG}$.
Step 5: Therefore $\frac{AF}{FB} = \frac{AE}{EC} \implies EF \parallel BC$ by converse of BPT.
Step 6: Since $EF \parallel BC$, $\triangle AEF \sim \triangle ABC$ by AA similarity. (Proved)
Final Answer: (i) OBGC is a ||gm, (ii) EF ∥ BC, (iii) ΔAEF ~ ΔABC (Proved)
Qsim18
2026
00:00
As shown in the given figure, a girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Step 1: Distance walked by girl in $4\text{ s} = 1.2 \times 4 = 4.8\text{ m}$.
Step 2: Height of lamp $= 3.6\text{ m}$, girl height $= 90\text{ cm} = 0.9\text{ m}$.
Step 3: Let shadow length $= x\text{ m}$. By similar triangles:
$$\frac{x}{x + 4.8} = \frac{0.9}{3.6} = \frac{1}{4}$$
Step 4: $4x = x + 4.8 \implies 3x = 4.8 \implies x = 1.6\text{ m}$.
Final Answer: 1.6 m
Qsim19
2026
00:00
Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) AQ = QR
(ii) AP = 2 PQ
(iii) PR = 2 AP
(i) AQ = QR
(ii) AP = 2 PQ
(iii) PR = 2 AP
Step 1: In $\triangle ADQ$ and $\triangle RCQ$: $DQ = QC$ (Q is midpoint), $\angle ADQ = \angle RCQ$ (alternate angles), $\angle AQD = \angle RQC$ (vertically opposite).
Step 2: By ASA, $\triangle ADQ \cong \triangle RCQ \implies AQ = QR$ and $AD = CR$.
Step 3: Since $AD = BC$, total length $BR = BC + CR = 2 AD$.
Step 4: In $\triangle APD$ and $\triangle RPB$: $AD \parallel BR \implies \triangle APD \sim \triangle RPB$.
Step 5: $\frac{AP}{PR} = \frac{AD}{BR} = \frac{AD}{2 AD} = \frac{1}{2} \implies PR = 2 AP$.
Step 6: Since $AR = AP + PR = 3 AP$ and $AR = 2 AQ$, we get $AP = 2 PQ$. (Proved)
Final Answer: (i) AQ=QR, (ii) AP=2PQ, (iii) PR=2AP (Proved)
Qsim20
2025C
00:00
In the figure, MNOP is a trapezium with MN ∥ PO and PO = 2 MN. A line segment FE drawn parallel to MN intersects MP at F and NO at E such that NE/EO = 3/4. Diagonal PN intersects FE at X. Prove that 7 FE = 10 MN.
Step 1: In $\triangle PNO$, $XE \parallel PO \implies \frac{NE}{NO} = \frac{3}{3+4} = \frac{3}{7}$.
Step 2: By similarity, $\frac{XE}{PO} = \frac{NE}{NO} = \frac{3}{7} \implies XE = \frac{3}{7} PO$.
Step 3: In $\triangle PMN$, $FX \parallel MN \implies \frac{FX}{MN} = \frac{PX}{PN} = \frac{OE}{ON} = \frac{4}{7} \implies FX = \frac{4}{7} MN$.
Step 4: Total $FE = FX + XE = \frac{4}{7} MN + \frac{3}{7} (2 MN) = \frac{10}{7} MN$.
Step 5: Cross-multiply: $7 FE = 10 MN$. (Proved)
Final Answer: 7 FE = 10 MN (Proved)
Qsim21
2025, 2024
00:00
In the given figure, PA, QB and RC are perpendicular to AC. If PA = x units, QB = y units and RC = z units, prove that 1/x + 1/z = 1/y.
Step 1: Since $PA, QB, RC \perp AC$, lines $PA, QB, RC$ are parallel to each other.
Step 2: In $\triangle PAC$, $QB \parallel PA \implies \frac{QB}{PA} = \frac{BC}{AC} \implies \frac{y}{x} = \frac{BC}{AC}$.
Step 3: In $\triangle RAC$, $QB \parallel RC \implies \frac{QB}{RC} = \frac{AB}{AC} \implies \frac{y}{z} = \frac{AB}{AC}$.
Step 4: Add both: $\frac{y}{x} + \frac{y}{z} = \frac{BC + AB}{AC} = \frac{AC}{AC} = 1$.
Step 5: Divide by $y$: $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$. (Proved)
Final Answer: 1/x + 1/z = 1/y (Proved)
Qsim22
2025
00:00
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR. Show that ΔABC ~ ΔPQR.
Step 1: Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
Step 2: Since $AD$ and $PM$ are medians, $BD = \frac{1}{2} BC$ and $QM = \frac{1}{2} QR$.
Step 3: Thus $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \implies \triangle ABD \sim \triangle PQM$ (by SSS similarity).
Step 4: Hence $\angle B = \angle Q$.
Step 5: In $\triangle ABC$ and $\triangle PQR$, $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q \implies \triangle ABC \sim \triangle PQR$ (by SAS). (Proved)
Final Answer: ΔABC ~ ΔPQR (Proved)
Qsim23
2025
00:00
The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA.
Step 1: In $\triangle ADF$ and $\triangle EBF$: $AD \parallel BC \implies \angle ADF = \angle EBF$ (alternate angles).
Step 2: $\angle AFD = \angle EFB$ (vertically opposite angles).
Step 3: By AA similarity, $\triangle ADF \sim \triangle EBF$.
Step 4: Therefore, $\frac{DF}{FB} = \frac{FA}{EF} \implies DF \times EF = FB \times FA$. (Proved)
Final Answer: DF × EF = FB × FA (Proved)
Qsim24
2025
00:00
In the adjoining figure, ΔCAB is a right triangle, right angled at A and AD ⊥ BC. Prove that ΔADB ~ ΔCDA. Further, if BC = 10 cm and CD = 2 cm, find the length of AD.
Step 1: In right $\triangle CAB$, $AD \perp BC \implies \angle BAD = \angle ACD$ and $\angle ABD = \angle CAD$.
Step 2: Thus $\triangle ADB \sim \triangle CDA$ (by AA similarity).
Step 3: Ratio of sides: $\frac{AD}{CD} = \frac{BD}{AD} \implies AD^2 = BD \times CD$.
Step 4: Here $BD = BC - CD = 10 - 2 = 8\text{ cm}$.
Step 5: $AD^2 = 8 \times 2 = 16 \implies AD = 4\text{ cm}$.
Final Answer: AD = 4 cm
Qsim25
2024C
00:00
In the given figure, MNOP is a parallelogram and AB ∥ MP. Prove that QC ∥ PO.
Step 1: In parallelogram $MNOP$, $MP \parallel NO$ and $MN \parallel PO$.
Step 2: Since $AB \parallel MP$, by similarity and BPT in the respective triangles, ratios of corresponding intercepts are equal.
Step 3: Thus $\frac{NQ}{QP} = \frac{NC}{CO} \implies QC \parallel PO$ by converse of BPT. (Proved)
Final Answer: QC ∥ PO (Proved)
Qsim26
2024
00:00
In the given figure, ΔFEC ≅ ΔGDB and ∠1 = ∠2. Prove that ΔADE ~ ΔABC.
Step 1: In $\triangle ADE$, $\angle 1 = \angle 2 \implies AD = AE$.
Step 2: Given $\triangle FEC \cong \triangle GDB \implies EC = DB$ (CPCT).
Step 3: Divide: $\frac{AD}{DB} = \frac{AE}{EC} \implies DE \parallel BC$ (by converse of BPT).
Step 4: Since $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$ by AA similarity. (Proved)
Final Answer: ΔADE ~ ΔABC (Proved)
Qsim27
2024, 2023
00:00
Sides 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.
Step 1: Extend $AD$ to $E$ such that $AD = DE$ and join $CE$. Extend $PM$ to $L$ such that $PM = ML$ and join $RL$.
Step 2: Quadrilaterals $ABEC$ and $PQLR$ are parallelograms because diagonals bisect each other.
Step 3: $\triangle ABE \sim \triangle PQL \implies \angle BAD = \angle QPM$.
Step 4: Similarly $\angle CAD = \angle RPM \implies \angle A = \angle P$.
Step 5: By SAS similarity, $\triangle ABC \sim \triangle PQR$. (Proved)
Final Answer: ΔABC ~ ΔPQR (Proved)
Qsim28
2023
00:00
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that EL = 2 BL.
Step 1: In $\triangle BMC$ and $\triangle EMD$: $MC = MD$, $\angle BMC = \angle EMD$, $\angle BCM = \angle EDM$ (alternate angles).
Step 2: $\triangle BMC \cong \triangle EMD \implies BC = ED$.
Step 3: Total length $AE = AD + DE = BC + BC = 2 BC$.
Step 4: In $\triangle AEL$ and $\triangle CBL$: $AE \parallel BC \implies \triangle AEL \sim \triangle CBL$.
Step 5: $\frac{EL}{BL} = \frac{AE}{BC} = \frac{2 BC}{BC} = 2 \implies EL = 2 BL$. (Proved)
Final Answer: EL = 2 BL (Proved)
Qsim29
2023
00:00
In the given figure, ΔABC and ΔDBC are on the same base BC. If AD intersects BC at O, prove that ar(ΔABC)/ar(ΔDBC) = AO/DO.
Step 1: Draw $AM \perp BC$ and $DN \perp BC$.
Step 2: In $\triangle AMO$ and $\triangle DNO$: $\angle AMO = \angle DNO = 90^\circ$, $\angle AOM = \angle DON$ (vertically opposite).
Step 3: $\triangle AMO \sim \triangle DNO \implies \frac{AM}{DN} = \frac{AO}{DO}$.
Step 4: $\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DBC)} = \frac{\frac{1}{2} \cdot BC \cdot AM}{\frac{1}{2} \cdot BC \cdot DN} = \frac{AM}{DN} = \frac{AO}{DO}$. (Proved)
Final Answer: ar(ΔABC)/ar(ΔDBC) = AO/DO (Proved)
Qsim30
NCERT, AI 2019
00:00
In the given figure, E is a point on CB produced of an isosceles ΔABC, with side AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Step 1: In isosceles $\triangle ABC$, $AB = AC \implies \angle B = \angle C$.
Step 2: In $\triangle ABD$ and $\triangle ECF$: $\angle ADB = \angle EFC = 90^\circ$ and $\angle ABD = \angle ECF$.
Step 3: By AA similarity, $\triangle ABD \sim \triangle ECF$. (Proved)
Final Answer: ΔABD ~ ΔECF (Proved)
Qsim31
Board Term I, 2017
00:00
In the given figure, ABC is a triangle and GHED is a rectangle. BC = 12 cm, HE = 6 cm, FC = BF and altitude AF = 24 cm. Find the area of the rectangle.
Step 1: Height of rectangle $HE = DG = 6\text{ cm}$. Altitude $AF = 24\text{ cm}$.
Step 2: Height of upper triangle $\triangle AGH = AF - HE = 24 - 6 = 18\text{ cm}$.
Step 3: Since $GH \parallel BC$, $\triangle AGH \sim \triangle ABC \implies \frac{GH}{BC} = \frac{18}{24} = \frac{3}{4}$.
Step 4: $GH = \frac{3}{4} \times 12 = 9\text{ cm}$.
Step 5: Area of rectangle $GHED = GH \times HE = 9 \times 6 = 54\text{ cm}^2$.
Final Answer: 54 cm²
Qsim32
Board Term I, 2017
00:00
Two poles of height 'p' and 'q' metres are standing vertically on a level ground, 'a' metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by pq/(p+q).
Step 1: Let height of intersection be $h$. Let distance from foot of pole $p$ be $x$ and pole $q$ be $a-x$.
Step 2: By similarity with pole $p$: $\frac{h}{p} = \frac{a-x}{a} = 1 - \frac{x}{a}$.
Step 3: By similarity with pole $q$: $\frac{h}{q} = \frac{x}{a}$.
Step 4: Add equations: $\frac{h}{p} + \frac{h}{q} = 1 - \frac{x}{a} + \frac{x}{a} = 1$.
Step 5: $h \left(\frac{1}{p} + \frac{1}{q}\right) = 1 \implies h \left(\frac{p+q}{pq}\right) = 1 \implies h = \frac{pq}{p+q}$. (Proved)
Final Answer: h = pq/(p+q) (Proved)
Qsim33
2023-24
00:00
In ΔABC, DE ∥ AB. If AB = a, DE = x, BE = b and EC = c. Express x in terms of a, b and c.
Step 1: Since $DE \parallel AB$, $\triangle CDE \sim \triangle CAB$.
Step 2: $\frac{DE}{AB} = \frac{CE}{CB} \implies \frac{x}{a} = \frac{c}{b+c}$.
Step 3: $x = \frac{ac}{b+c}$.
Final Answer: ac/(b+c) (Option b)