Exercise 8.1 — BPT Problems

Problems based on basic proportionality theorem.

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Exercise 8.1 — Similar Triangles | Class 10 Maths

Exercise 8.1 from Chapter 8 — Similar Triangles — is one of the most proof-intensive exercises in Class 10 Mathematics for CBSE, Telangana, and Andhra Pradesh board students. All nine questions in this exercise are based on just two key theorems learnt in the introduction: the Basic Proportionality Theorem (BPT / Thales Theorem) and its Converse. The final question (Q9) is a ruler-and-compass construction.

Mastering these proofs is essential — the proof technique of applying BPT twice and linking the two equations using a common intermediate ratio appears in Q2, Q3, Q6, and Q7. Recognise this pattern and every proof in this exercise becomes straightforward.

BPT Proofs Converse BPT Isosceles Triangle Mid-Point Theorems Trapezium Diagonals Line Division Construction

💡 Core strategy for this exercise: Identify which sub-triangle contains each parallel line. Apply BPT (or Converse BPT) to that sub-triangle. If you need to combine two ratio equations, look for a shared intermediate ratio that connects them.

Question	Topic	Theorem Used	Type
Q1	△PQR is isosceles (PS/SQ = PT/TR, ∠PST = ∠PRQ)	Converse BPT + corresponding angles	Proof
Q2	LM ∥ CB, LN ∥ CD → AM/AB = AN/AD	BPT twice, link via AL/AC	Proof
Q3	DE ∥ AC, DF ∥ AE → BF/FE = BE/EC	BPT twice, link via BD/DA	Proof
Q4	Line through midpoint ∥ another side bisects third side	BPT + midpoint condition	Proof
Q5	Line joining midpoints of two sides ∥ third side	Converse BPT + midpoint condition	Proof
Q6	DE ∥ OQ, DF ∥ OR → EF ∥ QR	BPT twice, link via PD/DO	Proof
Q7	AB ∥ PQ, AC ∥ PR → BC ∥ QR	BPT twice, link via OA/AP	Proof
Q8	Trapezium ABCD, AB ∥ DC → AO/BO = CO/DO	BPT + auxiliary line construction	Proof
Q9	Divide 7.2 cm segment in ratio 5:3	BPT (construction basis)	Construction

Question 1 — Prove △PQR is an Isosceles Triangle

Question 1

In △PQR, ST is a line such that PS/SQ = PT/TR and also ∠PST = ∠PRQ. Prove that △PQR is an isosceles triangle.

GivenPS/SQ = PT/TR and ∠PST = ∠PRQ in △PQR

To Prove△PQR is isosceles, i.e. PQ = PR

△PQR with ST ∥ QR

Step-by-Step Proof

①Given: PS/SQ = PT/TR in △PQR

②By Converse of BPT, ST ∥ QR

③ST ∥ QR and PQ is a transversal

∴ ∠PST = ∠PQR (corresponding angles) ...(1)

④Given: ∠PST = ∠PRQ ...(2)

⑤From (1) and (2): ∠PQR = ∠PRQ

⑥In △PQR, ∠PQR = ∠PRQ

∴ PQ = PR (sides opp. equal angles are equal)

✅ Hence Proved: △PQR is an isosceles triangle (PQ = PR).

💡 Key insight: The Converse BPT lets us conclude ST ∥ QR from the given ratio. Once we have the parallel line, corresponding angles give us ∠PST = ∠PQR. Combined with the given ∠PST = ∠PRQ, we get two equal base angles — the hallmark of an isosceles triangle.

Question 2 — Prove AM/AB = AN/AD (Double BPT)

Question 2

In the given figure, LM ∥ CB and LN ∥ CD. Prove that AM/AB = AN/AD.

GivenLM ∥ CB in △ABC; LN ∥ CD in △ADC

To ProveAM/AB = AN/AD

Figure for Q2

L is the common point on AC

Proof

①In △ABC, LM ∥ CB

By BPT: AM/AB = AL/AC ...(1)

②In △ADC, LN ∥ CD

By BPT: AL/AC = AN/AD ...(2)

③From (1) and (2): AM/AB = AN/AD

✅ Hence Proved. The link between the two equations is the common ratio AL/AC.

Question 3 — Prove BF/FE = BE/EC (Double BPT)

Question 3

In the given figure, DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.

GivenDE ∥ AC in △ABC; DF ∥ AE in △ABE

To ProveBF/FE = BE/EC

△ABC with D on AB, E and F on BC

Proof

①In △ABE, DF ∥ AE

By BPT: BD/DA = BF/FE ...(1)

②In △ABC, DE ∥ AC

By BPT: BD/DA = BE/EC ...(2)

③From (1) and (2): BF/FE = BE/EC

✅ Hence Proved. The linking ratio is BD/DA, which appears in both equations.

📌 Notice the pattern: Q2 and Q3 both use the same technique — apply BPT to two different sub-triangles and find the shared intermediate ratio that joins them. In Q2 it was AL/AC; here it is BD/DA.

Question 4 — Mid-Point Theorem (Part 1): Parallel Line Bisects Third Side

Question 4

Prove that a line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

GivenIn △ABC, D is the mid-point of AB (so AD = DB). A line through D ∥ BC meets AC at E.

To ProveE is the mid-point of AC, i.e. AE = EC

D = midpoint of AB, DE ∥ BC

Prove: E is midpoint of AC

Proof

①D is the midpoint of AB ⟹ AD = DB

∴ AD/DB = 1 ...(1)

②DE ∥ BC (given), DE meets AC at E

By BPT: AD/DB = AE/EC ...(2)

③From (1) and (2): AE/EC = 1

∴ AE = EC ⟹ E is the midpoint of AC

✅ Hence Proved: DE bisects AC — it passes through the midpoint of AC.

Question 5 — Mid-Point Theorem (Part 2): Join of Midpoints is Parallel

Question 5

Prove that a line joining the mid-points of any two sides of a triangle is parallel to the third side.

GivenIn △ABC, D is the midpoint of AB (AD = DB), E is the midpoint of AC (AE = EC).

To ProveDE ∥ BC

Proof

①D is midpoint of AB ⟹ AD = DB ⟹ AD/DB = 1 ...(1)

②E is midpoint of AC ⟹ AE = EC ⟹ AE/EC = 1 ...(2)

③From (1) and (2): AD/DB = AE/EC

i.e. DE divides AB and AC in the same ratio

④By Converse of BPT: DE ∥ BC

✅ Hence Proved. The line joining midpoints D and E is parallel to BC.

📌 Q4 vs Q5 — notice the difference:

Q4 starts with a parallel line through a midpoint → uses BPT to prove the other side is bisected.
Q5 starts with two known midpoints → uses Converse BPT to prove the join is parallel.

Together these two questions prove the complete Midpoint Theorem of triangles.

Question 6 — Show EF ∥ QR (Double BPT with Intermediate Point)

Question 6

In the given figure, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

GivenIn △POQ, DE ∥ OQ; in △POR, DF ∥ OR. D is a point on PO; E on PQ; F on PR.

To ProveEF ∥ QR (in △PQR)

△PQR with O inside, D on PO

Proof

①In △POQ, DE ∥ OQ

By BPT: PE/EQ = PD/DO ...(1)

②In △POR, DF ∥ OR

By BPT: PD/DO = PF/FR ...(2)

③From (1) and (2): PE/EQ = PF/FR

i.e. EF divides PQ and PR in the same ratio

④By Converse of BPT in △PQR: EF ∥ QR

✅ Hence Proved: EF ∥ QR. Linking ratio: PD/DO.

Question 7 — Show BC ∥ QR (AB ∥ PQ and AC ∥ PR)

Question 7

In the adjacent figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

GivenA on OP, B on OQ, C on OR; AB ∥ PQ (in △OPQ); AC ∥ PR (in △OPR)

To ProveBC ∥ QR (in △OQR)

Proof

①In △OPQ, AB ∥ PQ

By BPT: OA/AP = OB/BQ ...(1)

②In △OPR, AC ∥ PR

By BPT: OA/AP = OC/CR ...(2)

③From (1) and (2): OB/BQ = OC/CR

i.e. BC divides OQ and OR in the same ratio

④By Converse of BPT in △OQR: BC ∥ QR

✅ Hence Proved: BC ∥ QR. Linking ratio: OA/AP.

💡 Q6 vs Q7 comparison: Both use the same double-BPT technique. The difference is structural — in Q6, D is on PO (an internal point); in Q7, A is directly on OP. The proof logic is identical.

Question 8 — Trapezium ABCD: Show AO/BO = CO/DO

Question 8

ABCD is a trapezium in which AB ∥ DC and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.

GivenTrapezium ABCD with AB ∥ DC; diagonals AC and BD intersect at O.

To ProveAO/BO = CO/DO

Trapezium ABCD; EF ∥ AB through O

Construction + Proof

①Draw line EF through O, parallel to AB (and DC)(construction)

②In △ACD, EO ∥ DC(since EF ∥ DC)

By BPT: AO/CO = AE/DE ...(1)

③In △ABD, EO ∥ AB(since EF ∥ AB)

By BPT: AE/DE = BO/DO ...(2)

④From (1) and (2): AO/CO = BO/DO

⟹ AO/BO = CO/DO(rearranging — cross multiply)

✅ Hence Proved: AO/BO = CO/DO.

⚠️ Common mistake: Students forget to draw the auxiliary line EF through O. Without this construction, there is no triangle in which to apply BPT. The construction is a required step — always mention it in your proof.

Question 9 — Construction: Divide 7.2 cm in the Ratio 5 : 3

Question 9

Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.

Construction diagram — 7.2 cm divided in 5:3

AP = 4.5 cm | PB = 2.7 cm

📐 Steps of Construction

Draw a line segment AB = 7.2 cm.
Draw a ray AX making an acute angle with AB.
Mark off 5 + 3 = 8 equal parts on AX (label them A₁, A₂, …, A₈) using the same compass radius.
Join A₈ to B.
Draw a line through A₅ parallel to A₈B; let it meet AB at point P.
P divides AB in the ratio 5 : 3.
Measure with a ruler: AP = 4.5 cm and PB = 2.7 cm.

Verification by calculation: Total = 7.2 cm, ratio 5:3, so each part = 7.2 ÷ 8 = 0.9 cm AP = 5 × 0.9 = 4.5 cm PB = 3 × 0.9 = 2.7 cm Check: AP + PB = 4.5 + 2.7 = 7.2 cm ✓ Check: AP/PB = 4.5/2.7 = 5/3 ✓

💡 Why does this work? The construction uses the converse of BPT. A₅A₈ acts as the reference line, and the parallel through A₅ divides AB in the ratio 5:3 — exactly mirroring how the parallel line through A₅ divides the ray proportionally.

Common Mistakes to Avoid in Exercise 8.1

Applying BPT to the wrong triangle: Always name the specific sub-triangle you are applying BPT to (e.g. "In △ABE, DF ∥ AE"). Never just write "By BPT" without specifying the triangle — in board exams this loses marks.
Wrong ratio direction: BPT gives AD/DB = AE/EC (small segment/remaining segment). The alternate form is AD/AB = AE/AC. Don't mix these up mid-proof.
Forgetting the auxiliary line in Q8: The trapezium proof requires drawing EF ∥ AB through O. Omitting this construction means there is no triangle to apply BPT to — the proof collapses.
Confusing Q4 and Q5: Q4 uses direct BPT (parallel line given → prove bisection). Q5 uses Converse BPT (midpoints given → prove parallel). Know which direction you're working in.
Construction Q9 — wrong number of parts: For ratio 5:3, mark 5+3 = 8 parts, not 5 or 3 separately. Students sometimes mark only 5 and forget to add the denominator.

⛔ Board Exam Alert (Telangana & AP SSC): Q1 (isosceles triangle proof) and Q8 (trapezium diagonals) are the most commonly asked 4-mark proof questions from Exercise 8.1. For Q8, explicitly state: "Draw EF through O, parallel to AB and DC" as a Construction step before starting the proof — just like in the textbook solution.

Quick Reference — All Results at a Glance

Q#	What to Prove	Linking Ratio	Final Conclusion
Q1	△PQR is isosceles	∠PST = ∠PQR (corr. angles)	PQ = PR
Q2	AM/AB = AN/AD	AL/AC	Via BPT in △ABC and △ADC
Q3	BF/FE = BE/EC	BD/DA	Via BPT in △ABE and △ABC
Q4	Line through midpoint ∥ side → bisects 3rd side	AD/DB = 1	AE/EC = 1 (by BPT)
Q5	Join of midpoints ∥ 3rd side	AD/DB = AE/EC = 1	DE ∥ BC (Converse BPT)
Q6	EF ∥ QR	PD/DO	Via BPT in △POQ and △POR
Q7	BC ∥ QR	OA/AP	Via BPT in △OPQ and △OPR
Q8	AO/BO = CO/DO	AE/DE (auxiliary EF)	Via BPT in △ACD and △ABD
Q9	Divide 7.2 cm in 5:3	—	AP = 4.5 cm, PB = 2.7 cm

What Exercise 8.1 Prepares You For

The ratio-manipulation and parallel-line reasoning practised throughout Exercise 8.1 directly underlies the similarity criteria that follow — particularly the AA, SSS, and SAS similarity criteria in Exercise 8.2. The midpoint theorems from Q4 and Q5 also reappear in coordinate geometry proofs. The trapezium result from Q8 is a standard sub-step in several advanced similarity problems.

For background revision, see the Introduction to Similar Triangles which covers the full proof of BPT and the Converse. For connecting ideas from earlier classes, revisit Triangle Congruence from Class 9.

📐 Exam Tips — Telangana & AP SSC Board:

Proofs in Q1, Q8 are standard 4-mark exam questions — write the Given, To Prove, Construction (where applicable), and Proof sections clearly.
For Q4 and Q5, a neat labelled diagram with midpoint tick-marks earns you presentation marks.
Q9 (construction) carries 4 marks — each step of construction must be numbered and mentioned. Simply drawing the figure without steps will not get full marks.
When applying BPT, always write the name of the triangle first: "In △ABC, DE ∥ BC, by BPT, AD/DB = AE/EC" — not just the ratio alone.