Exercise 7.3 — More Congruence Criteria
More criteria for congruence of triangles.
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The Two Congruence Rules in This Exercise
Exercise 7.3 focuses on two powerful congruence rules that students must master for board exams: the SSS (Side-Side-Side) rule and the RHS (Right angle-Hypotenuse-Side) rule. These rules allow you to prove that two triangles are identical in shape and size — even without measuring angles.
SSS Rule
If all three sides of one triangle equal the three corresponding sides of another triangle, the triangles are congruent.
AB = PQ, BC = QR, AC = PR → △ABC ≅ △PQR
RHS Rule
In two right-angled triangles, if the hypotenuse and one side are equal, the triangles are congruent.
∠B = ∠Q = 90°, Hyp AC = PR, Side BC = QR → △ABC ≅ △PQR
💡 Remember CPCT: Once congruence is established, all corresponding parts — sides and angles — are equal. This is stated as CPCT (Corresponding Parts of Congruent Triangles). Every answer in this exercise uses CPCT as its final step.
Question 1 — Altitude of an Isosceles Triangle
Question 1
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that: (i) AD bisects BC (ii) AD bisects ∠A.
| Given | △ABC with AB = AC; AD ⊥ BC (AD is the altitude) |
| To Prove | (i) BD = CD (ii) ∠BAD = ∠CAD |
| Rule Used | RHS Congruence |
AD ⊥ BC, AB = AC
Solution
Consider triangles △ADB and △ADC:
1∠ADB = ∠ADC = 90° (AD ⊥ BC, given)
2AB = AC (given, hypotenuse)
3AD = AD (common side)
∴By RHS Rule: △ADB ≅ △ADC
Part (i): Since △ADB ≅ △ADC, by CPCT → BD = CD
∴ AD bisects BC ✓
Part (ii): Since △ADB ≅ △ADC, by CPCT → ∠BAD = ∠CAD
∴ AD bisects ∠A ✓
🎯 Key insight: The altitude of an isosceles triangle from the apex is also the perpendicular bisector of the base and the angle bisector — three properties in one line!
Question 2 — Two Sides and a Median Equal
Question 2
✓ Proved
Two sides AB, BC and median AM of △ABC are respectively equal to sides PQ, QR and median PN of △PQR. Show that: (i) △ABM ≅ △PQN (ii) △ABC ≅ △PQR.
| Given | AB = PQ, BC = QR, AM = PN (medians). M and N are midpoints of BC and QR respectively. |
| To Prove | (i) △ABM ≅ △PQN (ii) △ABC ≅ △PQR |
| Rules Used | SSS (Part i) → SAS (Part ii) |
△ABC with median AM
△PQR with median PN
Part (i): △ABM ≅ △PQN
Since M is the midpoint of BC: BM = ½ BC. Since N is the midpoint of QR: QN = ½ QR. Because BC = QR (given), we get BM = QN.
1AB = PQ (given)
2AM = PN (given — medians)
3BM = QN (½ BC = ½ QR)
∴By SSS Rule: △ABM ≅ △PQN
Part (ii): △ABC ≅ △PQR
From the congruence above, by CPCT: ∠ABM = ∠PQN (i.e., ∠ABC = ∠PQR). Now consider the full triangles:
1AB = PQ (given)
2BC = QR (given)
3∠ABC = ∠PQR (CPCT from part i)
∴By SAS Rule: △ABC ≅ △PQR
💡 Strategy tip: This is a two-step proof. You prove the smaller triangle pair first using SSS, extract an angle from CPCT, then use that angle to prove the bigger triangle pair with SAS. This "chain of congruence" technique is very common in board exams.
Question 3 — Equal Altitudes Prove Isosceles
Question 3
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that △ABC is isosceles.
| Given | BE ⊥ AC, CF ⊥ AB, BE = CF |
| To Prove | △ABC is isosceles, i.e., AB = AC |
| Rule Used | RHS Congruence |
BE = CF (equal altitudes)
Solution
We compare △BEC and △CFB to extract information about ∠B and ∠C:
1∠BEC = ∠CFB = 90° (altitudes, given)
2BC = CB (common hypotenuse)
3BE = CF (given — equal altitudes)
∴By RHS Rule: △BEC ≅ △CFB
By CPCT: ∠ECB = ∠FBC → ∠C = ∠B
In △ABC, sides opposite to equal angles are equal
∠C = ∠B → AB = AC
∴ △ABC is isosceles ✓
✓ Proved
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Question 4 — Base Angles of an Isosceles Triangle
Question 4
△ABC is an isosceles triangle in which AB = AC. Show that ∠B = ∠C. (Hint: Draw AP ⊥ BC; use RHS rule.)
| Given | △ABC with AB = AC |
| Construction | Draw AP ⊥ BC, so ∠APB = ∠APC = 90° |
| To Prove | ∠B = ∠C |
| Rule Used | RHS Congruence |
Construction: AP ⊥ BC
Solution
1∠APB = ∠APC = 90° (by construction)
2AB = AC (given — hypotenuse)
3AP = AP (common side)
∴By RHS Rule: △APB ≅ △APC
By CPCT: ∠B = ∠C ✓
Theorem (Isosceles Triangle Theorem)
The base angles of an isosceles triangle are always equal. This theorem is one of the most frequently tested results in Class 9 board exams for both CBSE and the Telangana/AP syllabi.
Question 5 — Proving ∠BCD is a Right Angle
Question 5
△ABC is isosceles (AB = AC). Side BA is extended to D such that AD = AB. Show that ∠BCD = 90°.
| Given | AB = AC (△ABC isosceles); D on BA extended, AD = AB, so AD = AC. |
| To Prove | ∠BCD = 90° |
| Key Idea | Use exterior angle theorem and linear pair property |
D on BA extended; AD = AB = AC
Solution — Step by Step
In △ABC: AB = AC → ∠ABC = ∠ACB = x° (base angles of isosceles △)
In △ADC: AD = AC (since AD = AB = AC) → ∠ADC = ∠ACD = y°
Exterior angle of △ABC at A:
∠DAC = ∠ABC + ∠ACB = x° + x° = 2x° … (1)
Exterior angle of △ADC at A:
∠BAC = ∠ADC + ∠ACD = y° + y° = 2y° … (2)
∠DAC + ∠BAC = 180° (linear pair on straight line BD)
2x° + 2y° = 180°
2(x° + y°) = 180°
x° + y° = 90°
∠BCD = ∠ACB + ∠ACD = x° + y° = 90° ✓
✓ ∠BCD = 90° Proved
💡 Exterior Angle Theorem reminder: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles. This property is the engine that drives Question 5's proof.
Question 6 — Right-Isosceles Triangle: ∠B = ∠C
Question 6
△ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Show that ∠B = ∠C.
| Given | ∠A = 90°, AB = AC |
| Construction | Draw AD ⊥ BC |
| To Prove | ∠B = ∠C |
| Rule Used | RHS Congruence |
∠A = 90°, AB = AC; AD ⊥ BC drawn
1∠ADB = ∠ADC = 90° (construction)
2AB = AC (given — equal hypotenuses)
3AD = AD (common)
∴By RHS: △ADB ≅ △ADC → ∠B = ∠C ✓
📌 This is an isosceles right triangle. Since ∠A = 90°, the remaining angles are ∠B = ∠C = 45°. The proof doesn't require you to calculate the angles — just show they are equal.
Question 7 — All Angles of an Equilateral Triangle = 60°
Question 7
Show that the angles of an equilateral triangle are 60° each.
Equilateral △: AB = BC = CA
Proof using the Isosceles Triangle Theorem:
In △ABC, AB = BC = CA
AB = AC → ∠B = ∠C … (1) [base angles of isosceles △]
BC = CA → ∠B = ∠A … (2) [base angles of isosceles △]
From (1) and (2): ∠A = ∠B = ∠C
∠A + ∠B + ∠C = 180° (angle sum property)
3∠A = 180°
∠A = ∠B = ∠C = 60° ✓
✓ Each angle = 60°
Question 8 — Extensions of Isosceles Sides
Question 8
△ABC is isosceles (AB = AC). BA and CA are extended to Q and P respectively such that AQ = AP. Show that PB = QC.
| Given | AB = AC; BA extended to Q, CA extended to P; AQ = AP |
| To Prove | PB = QC |
| Key Angle | ∠PAB = ∠QAC (vertically opposite angles) |
| Rule Used | SAS Congruence |
BA → Q, CA → P; AQ = AP
Solution
Compare △ABP and △ACQ:
1AB = AC (given)
2∠PAB = ∠QAC (vertically opposite angles)
3AP = AQ (given)
∴By SAS Rule: △ABP ≅ △ACQ
By CPCT: PB = QC ✓
✓ PB = QC Proved
Quick Summary — All 8 Questions at a Glance
| Q | What is Given | What to Prove / Show | Rule Used |
|---|---|---|---|
| Q1(i) | Isosceles △ABC, AD altitude | AD bisects BC | RHS + CPCT |
| Q1(ii) | Isosceles △ABC, AD altitude | AD bisects ∠A | RHS + CPCT |
| Q2(i) | AB=PQ, BC=QR, median AM=PN | △ABM ≅ △PQN | SSS |
| Q2(ii) | From Q2(i) | △ABC ≅ △PQR | SAS |
| Q3 | Equal altitudes BE = CF | △ABC is isosceles | RHS + CPCT |
| Q4 | Isosceles △ABC, AB = AC | ∠B = ∠C | RHS + CPCT |
| Q5 | Isosceles △ABC; AD = AB, D on BA extended | ∠BCD = 90° | Exterior Angle + Linear Pair |
| Q6 | ∠A = 90°, AB = AC in △ABC | ∠B = ∠C | RHS + CPCT |
| Q7 | Equilateral △ABC | Each angle = 60° | Isosceles + Angle Sum |
| Q8 | AB=AC, AQ=AP, P and Q on extensions | PB = QC | SAS + CPCT |
Common Mistakes to Avoid in Board Exams
- Wrong order of congruence: Always write △ABC ≅ △PQR with vertices in the correct matching order. If you write △ABC ≅ △QPR, you are claiming a different correspondence and CPCT will give wrong results.
- Forgetting to state CPCT: Once you declare two triangles congruent, you must explicitly write "by CPCT" when extracting equal sides or angles. Skipping this step costs marks in Telangana and AP board evaluations.
- Mixing up the RHS conditions: The RHS rule needs (1) a right angle, (2) the hypotenuse equal, and (3) one other side equal — not two other sides. If you have two sides and a right angle that is not between them, use RHS not SAS.
- Confusing the altitude with the median: An altitude is perpendicular to the opposite side. A median connects to the midpoint. In an isosceles triangle, the altitude from the apex doubles as the median — but this must be proved, not assumed.
- Not mentioning construction in Q4 and Q6: When you draw an auxiliary line like AP ⊥ BC, state it as a construction step before the proof. Examiners expect it explicitly.
⛔ Most common board exam error: Writing ∠BAD = ∠CAD without mentioning that this follows "by CPCT from △ADB ≅ △ADC". The congruence must always come before CPCT — never the reverse.
What This Exercise Prepares You For
The SSS and RHS congruence rules from Exercise 7.3 are foundational tools you will use throughout Class 9 and Class 10. The next exercise (7.4) covers inequalities in triangles, where you prove which side or angle is larger — a topic that builds directly on the congruence logic developed here.
The isosceles triangle results proved in Q1, Q4, and Q6 reappear in earlier exercises and are used as ready-made results in Class 10 geometry, especially in Chapter 6 — Triangles (similarity) and in circle theorems where isosceles triangles formed by equal radii are routine.
For Telangana and Andhra Pradesh board examinations, proofs from Chapter 7 typically appear as 4-mark or 5-mark questions. The key to full marks is a clean structure: Given → To Prove → Construction (if any) → Proof (with rule stated) → Conclusion with CPCT.
✅ Board Exam Tip (CBSE, Telangana & AP): For 4-mark proof questions from this exercise, always write a "Given" and "To Prove" table at the top. Examiners at all three boards award 1 mark simply for setting this up correctly, before even reading your proof.