Geometry of Triangles: NCERT Ex 7.3 Explained for Class 9

Understanding SSS and RHS Congruence Rules

Theorem 7.4: SSS (Side-Side-Side) Congruence Rule

This rule is one of the most intuitive. It states that if all three sides of one triangle are equal to the three corresponding sides of another triangle, then the two triangles must be congruent. Imagine you have three sticks of fixed lengths. No matter how you try, you can only form one specific triangle with them. This rigidity is the core idea behind SSS congruence. So, if ΔABC and ΔPQR have AB = PQ, BC = QR, and AC = PR, we can confidently state that ΔABC ≅ ΔPQR.

Theorem 7.5: RHS (Right-angle-Hypotenuse-Side) Congruence Rule

This is a special rule exclusively for right-angled triangles. It states that if the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one corresponding side of another right-angled triangle, then the two triangles are congruent. The name itself is a checklist:

• R - Both triangles must have a Right angle (90°).
• H - The Hypotenuses (the side opposite the right angle) must be equal.
• S - One pair of corresponding Sides (other than the hypotenuse) must be equal.

If all three conditions are met, you can use the RHS rule to prove congruence. This is different from SSA, which is not a valid rule!

How to Apply Congruence Rules in Proofs

1. Step 1: Analyze the Given Information & Diagram — Carefully read the problem and examine the diagram. List all the given facts, such as equal sides or angles. Clearly state what you need to 'Prove'. This is your starting point and your goal.
2. Step 2: Identify the Triangles and Choose a Rule — Identify the two triangles you need to prove congruent. Based on the 'Given' information, decide which congruence rule to use. Do you have three pairs of equal sides (SSS)? Or do you have two right-angled triangles with equal hypotenuses and one pair of equal sides (RHS)?
3. Step 3: Write the Proof Systematically — In the two triangles you've chosen (e.g., in ΔABC and ΔDEF), list the three pairs of equal parts. For each pair, write the reason in brackets, e.g., (Given), (Common side), (Proved above).
4. Step 4: State the Conclusion and Use CPCTC — After listing three valid pairs, conclude the proof by stating that the triangles are congruent and mentioning the rule used (e.g., 'Therefore, ΔABC ≅ ΔDEF by SSS rule'). If needed, you can then deduce that other parts are equal by using 'CPCTC' (Corresponding Parts of Congruent Triangles are Equal).

Worked Example: Applying the SSS Rule

• Problem: ΔABC and ΔDBC are two isosceles triangles on the same base BC such that vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that ΔABD ≅ ΔACD. Solution: Given: ΔABC is isosceles with AB = AC. ΔDBC is isosceles with DB = DC. To Prove: ΔABD ≅ ΔACD Proof: In ΔABD and ΔACD, 1. AB = AC (Given, since ΔABC is isosceles) 2. DB = DC (Given, since ΔDBC is isosceles) 3. AD = AD (Common side) Conclusion: Since all three corresponding sides are equal, by the SSS congruence rule, ΔABD ≅ ΔACD.

Exam Tip: Don't Confuse RHS with SSA!

A very common mistake is to try and prove triangles congruent using two sides and a non-included angle (SSA). Remember, SSA is NOT a valid congruence criterion! The RHS rule looks similar, but it has very specific requirements. For RHS to be valid, the angle must be a 90-degree right angle, and one of the equal sides must be the hypotenuse (the side opposite the right angle). If you have two sides and an angle that is not between them, and it's not a right-angled situation fitting RHS, you cannot claim the triangles are congruent. Always check: is it a right angle? Is the hypotenuse involved? If not, you cannot use RHS.

Practice Questions with Solutions

• Q: AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that ΔADB ≅ ΔADC. A: Step 1: Identify the given information and the triangles to be compared. We are given AB = AC and AD ⊥ BC, which means ∠ADB = ∠ADC = 90°. We need to prove ΔADB ≅ ΔADC. Step 2: Check for the appropriate congruence rule. We have two right-angled triangles. The hypotenuses are AB and AC, which are given as equal. The side AD is common to both triangles. This fits the RHS (Right-angle-Hypotenuse-Side) criteria. Step 3: Write down the proof. In right-angled ΔADB and ΔADC: - ∠ADB = ∠ADC (Each 90°, since AD is an altitude) - AB = AC (Given, hypotenuses) - AD = AD (Common side) Final answer: Therefore, by the RHS congruence rule, ΔADB ≅ ΔADC.
• Q: In ΔPQR, if PS is a median such that PS = QS = RS, what can you say about ∠QPR? A: Step 1: Analyze the given information. PS is a median, so S is the midpoint of QR. We are also given PS = QS = RS. This means S is equidistant from all three vertices P, Q, and R. Step 2: Consider ΔPSQ and ΔPSR. In ΔPSQ, since PS = QS, it is an isosceles triangle. Therefore, ∠SPQ = ∠SQP (angles opposite to equal sides). Let's call this angle x. So, ∠SPQ = ∠PQR = x. Step 3: Similarly, in ΔPSR, since PS = RS, it is also an isosceles triangle. Therefore, ∠SPR = ∠SRP. Let's call this angle y. So, ∠SPR = ∠PRQ = y. Step 4: Apply the angle sum property in the larger triangle, ΔPQR. We have ∠PQR + ∠PRQ + ∠QPR = 180°. We know ∠QPR = ∠SPQ + ∠SPR = x + y. Substituting the values: x + y + (x + y) = 180° is incorrect. The correct sum is (∠PQR) + (∠PRQ) + (∠QPR) = x + y + (x+y) = 180°. So, 2(x+y) = 180°. This means x+y = 90°. Final answer: Since ∠QPR = x + y, we have ∠QPR = 90°. The angle is a right angle.
• Q: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR. Show that ΔABM ≅ ΔPQN. A: Step 1: List the given equalities. We are given: AB = PQ, BC = QR, and median AM = median PN. Step 2: Use the property of medians. Since AM is the median to BC, M is the midpoint of BC. So, BM = 1/2 BC. Similarly, since PN is the median to QR, N is the midpoint of QR. So, QN = 1/2 QR. Step 3: Relate the side lengths. We are given BC = QR. Taking half on both sides gives 1/2 BC = 1/2 QR, which implies BM = QN. Step 4: Apply the SSS congruence rule to ΔABM and ΔPQN. In these two triangles: - AB = PQ (Given) - AM = PN (Given) - BM = QN (Proved in Step 3) Final answer: Since all three corresponding sides are equal, by the SSS congruence rule, ΔABM ≅ ΔPQN.
• Q: In a square ABCD, P is a point on side BC such that AP = DP. Prove that ΔABP ≅ ΔDCP. A: Step 1: Identify the properties of a square and the given information. In a square ABCD, all sides are equal (AB = BC = CD = DA) and all angles are 90° (∠B = ∠C = 90°). We are also given AP = DP. Step 2: Choose the triangles and the congruence rule. We need to prove ΔABP ≅ ΔDCP. These are right-angled triangles (at B and C). We are given their hypotenuses are equal (AP = DP). We also know a pair of corresponding sides are equal (AB = DC, sides of a square). This fits the RHS congruence rule. Step 3: Write down the proof. In right-angled ΔABP and ΔDCP: - ∠ABP = ∠DCP (Each 90°, angles of a square) - AP = DP (Given, hypotenuses) - AB = DC (Sides of a square) Final answer: Therefore, by the RHS congruence rule, ΔABP ≅ ΔDCP.

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