Triangles

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CLASS IX Mathematics ~8–10 marks/year Ch 7 of 12
Triangles

Class 9 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Congruence means two figures are identical in shape and size; the symbol is ≅ (congruent to). Superposing one on the other leaves no gap.
  • Five criteria guarantee triangle congruence without checking all six elements: SSS, SAS, ASA, AAS, RHS.
  • CPCT (Corresponding Parts of Congruent Triangles are equal) is the key conclusion after any congruence proof.
  • Isosceles triangle theorem — equal sides force equal opposite angles; the converse is also true.
  • Inequalities — the greater side has the greater opposite angle; the sum of any two sides exceeds the third side.
  • Board weightage: 8–10 marks/year — expect one 5-mark congruence proof, one 3-mark theorem proof, and 1–2 MCQs each year.
Detailed Notes

1. Congruence of Figures

Two figures are congruent if they have exactly the same shape and the same size. The standard test is superposition: if one figure placed exactly on top of the other covers it completely (with no overlap and no gap), the two figures are congruent.

  • Two line segments are congruent ⇔ they have equal length.
  • Two angles are congruent ⇔ they have equal measure.
  • Two circles are congruent ⇔ they have equal radii.
  • Two squares are congruent ⇔ they have equal sides.

Notation: AB ≅ CD means segment AB is congruent to segment CD, i.e., AB = CD in length. For triangles we write △ABC ≅ △PQR, which encodes the specific correspondence A↔P, B↔Q, C↔R.

Order matters. △ABC ≅ △PQR and △ABC ≅ △QPR are different statements. In the first, A corresponds to P; in the second, A corresponds to Q. Always write the vertices in the order of correspondence.

2. Congruence of Triangles and CPCT

Two triangles are congruent if all six corresponding parts (three sides and three angles) are equal. If △ABC ≅ △PQR, then:

  • AB = PQ,   BC = QR,   CA = RP   (three pairs of equal sides)
  • ∠A = ∠P,   ∠B = ∠Q,   ∠C = ∠R   (three pairs of equal angles)

CPCT — Corresponding Parts of Congruent Triangles are equal. This is the rule you invoke after proving congruence. You never write “AB = PQ by CPCT” before establishing congruence; the congruence statement must come first.

Why congruence criteria? Verifying all six elements is impractical in a proof. The five criteria below each show that three specific elements are enough to guarantee full congruence. Learning which three elements are sufficient — and which combinations are NOT — is the core skill of this chapter.

NCERT Example 1 — first CPCT application

In △ABC, the bisector AD of ∠A meets BC at D. If AB = AC, prove △ABD ≅ △ACD and hence BD = DC.

In △ABD and △ACD:
AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common).
By SAS: △ABD ≅ △ACD. Therefore BD = DC (CPCT).

3. SAS Congruence Criterion (Axiom 7.1)

In NCERT, SAS is taken as an axiom (accepted without proof) because it cannot be proved from simpler axioms alone. All other criteria are theorems proved using SAS.

Axiom 7.1 (SAS): Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.

If AB = DE, ∠B = ∠E, BC = EF (angle B is between sides AB and BC; angle E is between DE and EF), then △ABC ≅ △DEF.

Critical point — “included” is not optional. The equal angle must be sandwiched between the two equal sides. If the equal angle is not included (i.e., SSA), the criterion fails — two different triangles can satisfy SSA.

NCERT Example 2 — SAS in action

OA = OB and OC = OD (O is a point). Prove AC = BD.

In △AOC and △BOD: OA = OB (given), ∠AOC = ∠BOD (vertically opposite angles), OC = OD (given).
By SAS: △AOC ≅ △BOD. Therefore AC = BD (CPCT). ■

Exercise 7.1 Q1 — SAS with altitudes

In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show △ABC ≅ △ABD and conclude BC = BD.

In △ABC and △ABD: AC = AD (given), ∠CAB = ∠DAB (AB bisects ∠A), AB = AB (common).
By SAS: △ABC ≅ △ABD. Therefore BC = BD and ∠ABC = ∠ABD (CPCT).

4. ASA Congruence Criterion (Theorem 7.1)

Theorem 7.1 (ASA): Two triangles are congruent if two angles and the included side of one triangle are equal to the two angles and the included side of the other triangle.

If ∠B = ∠E, BC = EF, ∠C = ∠F, then △ABC ≅ △DEF.

Full NCERT Proof:

Given △ABC and △DEF with ∠B = ∠E, BC = EF, ∠C = ∠F. We wish to prove △ABC ≅ △DEF.

Step 1: Place △ABC on △DEF so that B falls on E and BC falls along EF. Since BC = EF, vertex C falls exactly on F.

Step 2: Ray BA makes angle ∠B with BC. Since ∠B = ∠E, ray BA falls along ray ED. So A lies somewhere on ray ED.

Step 3: Ray CA makes angle ∠C with CB. Since ∠C = ∠F, ray CA falls along ray FD. So A lies somewhere on ray FD.

Step 4: A lies on both ray ED and ray FD. Two non-parallel, non-coincident lines intersect at exactly one point. That point is D. Hence A coincides with D.

With all three vertices coinciding, △ABC ≅ △DEF. ■

NCERT Example 3 — ASA application

Line segment AB is bisected by PQ at M, so AM = BM. ∠PAM = ∠QBM and ∠APM = ∠BQM. Prove that AP = BQ.

In △APM and △BQM: ∠APM = ∠BQM (given), AM = BM (M bisects AB), ∠PAM = ∠QBM (given).
By ASA: △APM ≅ △BQM. Therefore AP = BQ (CPCT). ■

5. AAS Congruence Criterion (Theorem 7.2)

Theorem 7.2 (AAS): Two triangles are congruent if two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle.

If ∠A = ∠D, ∠B = ∠E, BC = EF (BC is opposite ∠A; EF is opposite ∠D — not necessarily the included side), then △ABC ≅ △DEF.

Why AAS follows from ASA: Since ∠A = ∠D and ∠B = ∠E, by the angle-sum property:
∠C = 180° − ∠A − ∠B = 180° − ∠D − ∠E = ∠F.
So ∠B = ∠E, BC = EF, ∠C = ∠F — this is exactly ASA. Hence △ABC ≅ △DEF by ASA ⇒ AAS is proved.

NCERT Example 4 — AAS with perpendicular bisector

In △ABC, the perpendicular bisector of BC passes through A. Prove AB = AC.

Let D be the midpoint of BC and AD ⊥ BC. In △ADB and △ADC:
∠ADB = ∠ADC = 90° (AD ⊥ BC), BD = CD (D midpoint), AD = AD (common).
By SAS: △ADB ≅ △ADC. Therefore AB = AC (CPCT) — △ABC is isosceles. ■

Exercise 7.1 Q4 — equal altitudes

BE and CF are altitudes of △ABC with BE = CF. Prove △ABE ≅ △ACF and deduce AB = AC.

In △ABE and △ACF: ∠AEB = ∠AFC = 90°, ∠A = ∠A (common), BE = CF (given).
By AAS: △ABE ≅ △ACF. Therefore AB = AC (CPCT). ■

6. SSS Congruence Criterion (Theorem 7.4)

Theorem 7.4 (SSS): If three sides of one triangle are equal to the three sides of another triangle, the two triangles are congruent.

If AB = DE, BC = EF, CA = FD, then △ABC ≅ △DEF.

Proof idea: Place △ABC on △DEF with B on E, BC along EF. Since BC = EF, C reaches F. Now AB = DE so A is at distance DE from E; CA = FD so A is also at distance FD from F. By construction (using the intersection of two arcs), A must coincide with D. Hence all three vertices coincide and the triangles are congruent. ■

NCERT Exercise 7.3 Q3 — SSS

Two sides AB, BC and median AM of one triangle are respectively equal to sides PQ, QR and median PN of another triangle. Show △ABC ≅ △PQR.

M is midpoint of BC, N midpoint of QR, so BM = QN (half of equal sides). In △ABM and △PQN: AB = PQ (given), BM = QN (half of BC = QR), AM = PN (given medians). By SSS: △ABM ≅ △PQN. Therefore ∠B = ∠Q (CPCT). Now in △ABC and △PQR: AB = PQ, ∠B = ∠Q, BC = QR. By SAS: △ABC ≅ △PQR. ■

7. RHS Congruence Criterion (Theorem 7.5)

Theorem 7.5 (RHS): If in two right triangles the hypotenuse and one side of one are equal to the hypotenuse and one side of the other, the two triangles are congruent.

If ∠B = ∠E = 90°, AC = DF (hypotenuses), AB = DE (one leg), then △ABC ≅ △DEF.

Proof — why RHS works: By Pythagoras’ theorem in △ABC: BC² = AC² − AB². In △DEF: EF² = DF² − DE². Since AC = DF and AB = DE, we get BC² = EF², so BC = EF. Now AB = DE, BC = EF, ∠B = ∠E = 90° — this is SAS. Hence △ABC ≅ △DEF. ■

Key restriction: RHS applies only to right-angled triangles. The “R” stands for the right angle (90°) present in both triangles.

NCERT Example 5 — RHS

P is a point equidistant from lines l and m which intersect at A. Show that AP bisects the angle between l and m.

Draw PB ⊥ l and PC ⊥ m. PB = PC (given equal distances). In △PAB and △PAC: ∠PBA = ∠PCA = 90°, PA = PA (common hypotenuse), PB = PC (given). By RHS: △PAB ≅ △PAC. Therefore ∠PAB = ∠PAC (CPCT), so AP bisects angle A. ■

Exercise 7.2 Q4 — RHS

ABC is an isosceles triangle with AB = AC. The side BA is produced to D such that AD = AB. Prove that ∠BCD is a right angle.

AB = AC and AD = AB, so AC = AD. In △ABC: ∠ACB = ∠ABC (isosceles, AB = AC). Let ∠ACB = ∠ABC = x. In △ACD: AC = AD so ∠ACD = ∠ADC = y. Now ∠BCD = ∠ACB + ∠ACD = x + y. In △BCD: ∠BDC + ∠DBC + ∠BCD = 180°. Angles y + 2x + (x+y) = 180° ... Using ∠A = 180 − 2x in triangle ABC and exterior angle theorem, the final conclusion is ∠BCD = 90°. ■

8. Isosceles Triangle Theorem — Equal Sides Give Equal Angles

Theorem 7.2 (NCERT): Angles opposite to equal sides of an isosceles triangle are equal.

If △ABC has AB = AC, then ∠ABC = ∠ACB.

Full Proof:

Construct the bisector AD of ∠BAC, where D is the point where the bisector meets BC.

In △ABD and △ACD:

  • AB = AC   (given)
  • ∠BAD = ∠CAD   (AD bisects ∠BAC)
  • AD = AD   (common side)

By SAS: △ABD ≅ △ACD.

Therefore ∠ABD = ∠ACD (CPCT), i.e. ∠ABC = ∠ACB. ■

Corollary 1: Every equilateral triangle is also equiangular (all three angles = 60°).

Corollary 2: The bisector of the vertical angle of an isosceles triangle is also the perpendicular bisector of the base and the median to the base — all four special lines coincide at AD.

NCERT Example 6 — combining isosceles theorem with congruence

In △ABC, AB = BC (isosceles). The bisector of ∠B meets AC at D. Prove: BD ⊥ AC.

By Theorem 7.2, ∠BAC = ∠BCA (equal sides AB = BC). Let this common angle = x. In △ABD and △CBD: ∠BAD = ∠BCD = x, ∠ABD = ∠CBD (BD bisects ∠B), BD = BD (common). By AAS: △ABD ≅ △CBD. So ∠BDA = ∠BDC. Since ∠BDA + ∠BDC = 180° (straight line), each = 90°. Hence BD ⊥ AC. ■

9. Converse — Equal Angles Give Equal Sides (Theorem 7.3)

Theorem 7.3 (NCERT): The sides opposite to equal angles of a triangle are equal.

If in △ABC, ∠ABC = ∠ACB, then AB = AC.

Full Proof:

Construct the bisector AD of ∠BAC, D on BC. In △ABD and △ACD:

  • ∠ABD = ∠ACD   (given, these are ∠ABC and ∠ACB)
  • ∠BAD = ∠CAD   (AD bisects ∠A)
  • AD = AD   (common)

By AAS: △ABD ≅ △ACD. Therefore AB = AC (CPCT). ■

Corollary: Every equiangular triangle is equilateral.

NCERT Example 7 — applying the converse

E and F are points on equal sides AB and AC of isosceles △ABC (AB = AC) such that BE = CF. Prove △BCF ≅ △CBE and hence BF = CE.

Since AB = AC, by Theorem 7.2, ∠ABC = ∠ACB. Also AB − BE = AC − CF ⇒ AE = AF, but we need BC = BC. In △BCF and △CBE: BC = CB (common), ∠CBF = ∠BCE (i.e. ∠ABC = ∠ACB), CF = BE (given). By SAS: △BCF ≅ △CBE. Therefore BF = CE (CPCT). ■

Exercise 7.2 Q1

In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD. Show AD = AE.

In △ABD and △ACE: AB = AC (given), ∠B = ∠C (equal sides ⇒ equal angles), BD = CE (since BE = CD and BC = BD + DC = CE + EB; so BD = CE). By SAS: △ABD ≅ △ACE. Therefore AD = AE (CPCT). ■

10. Inequalities — Greater Side Has Greater Opposite Angle

Theorem 7.6 (NCERT): If two sides of a triangle are unequal, the angle opposite to the longer side is larger (greater).

In △ABC, if AB > AC, then ∠ACB > ∠ABC.

Proof: Since AB > AC, mark D on AB such that AD = AC. In △ACD: AD = AC means ∠ACD = ∠ADC (isosceles triangle theorem). Now ∠ACB = ∠ACD + ∠DCB > ∠ACD = ∠ADC. But ∠ADC is an exterior angle of △BCD, so ∠ADC > ∠DBC = ∠ABC. Combining: ∠ACB > ∠ABC. ■

Theorem 7.7 (Converse): In any triangle, the side opposite to the greater angle is longer.

In △ABC, if ∠ACB > ∠ABC, then AB > AC.

Together these mean: Largest angle ↔ longest side; smallest angle ↔ shortest side. This is the fundamental inequality linking angles and sides.

Finding the order of sides from angles

In △ABC, ∠A = 50°, ∠B = 70°, ∠C = 60°. Arrange sides in increasing order.

Largest angle = ∠B = 70° ⇒ opposite side AC is longest. Smallest angle = ∠A = 50° ⇒ opposite side BC is shortest. Middle = ∠C = 60° ⇒ side AB is middle. Order: BC < AB < AC.

NCERT Example 8 — inequality chain

D is a point on side BC of △ABC such that AD = AC. Show that AB > AD.

In △ADC, AD = AC means ∠ACD = ∠ADC (isosceles). ∠ADB is an exterior angle of △ADC, so ∠ADB > ∠ACD. In △ABC, ∠ABD = ∠ABC and from above ∠ADB > ∠ABC. In △ABD, ∠ADB > ∠ABD means AB > AD (side opposite greater angle). ■

11. Triangle Inequality — Sum of Two Sides > Third Side

Theorem 7.8 (NCERT): The sum of any two sides of a triangle is greater than the third side.

In any △ABC: AB + BC > CA,   BC + CA > AB,   CA + AB > BC.

Full Proof: Extend BA to D such that AD = AC (so BD = AB + AD = AB + AC). In △ACD, AC = AD means ∠ACD = ∠ADC. Now ∠BCD = ∠BCA + ∠ACD > ∠ACD = ∠ADC = ∠BDC. In △BCD, ∠BCD > ∠BDC ⇒ BD > BC (side opposite greater angle). But BD = AB + AC. So AB + AC > BC. ■

Useful corollary: The difference of any two sides is less than the third: |AB − AC| < BC. This gives a two-sided bound: |AB − AC| < BC < AB + AC.

Quick exam test: To decide if side lengths a, b, c form a valid triangle, check only: (smallest) + (middle) > largest. If a ≤ b ≤ c, check a + b > c. If that holds, the triangle exists.

Exercise 7.4 Q1 — right triangle inequality

Show that in a right-angled triangle, the hypotenuse is the longest side.

In △ABC with ∠B = 90°: ∠A + ∠C = 90°, so ∠A < 90° = ∠B and ∠C < 90° = ∠B. Thus ∠B is the greatest angle. The side opposite ∠B is AC (hypotenuse). By Theorem 7.7, AC > AB and AC > BC. The hypotenuse is the longest side. ■

Exercise 7.4 Q2 — median inequality

In △ABC, AM is a median. Prove AB + BC + CA > 2AM.

In △ABM: AB + BM > AM (triangle inequality). In △ACM: AC + CM > AM (triangle inequality). Adding: AB + BM + AC + CM > 2AM, i.e. AB + AC + BC > 2AM (since BM + CM = BC). ■

Exercise 7.4 Q4 — sum of medians

Show that the perimeter of a triangle is greater than the sum of its three medians.

For each median, the triangle-inequality gives: AB + BC > 2AM, BC + CA > 2BN, CA + AB > 2CL (where AM, BN, CL are medians — proved as above). Adding all three: 2(AB + BC + CA) > 2(AM + BN + CL), hence AB + BC + CA > AM + BN + CL. Perimeter > sum of medians. ■

12. Summary Table — All Five Congruence Criteria

CriterionElements neededKey restrictionType
SAS2 sides + included angleAngle must be between the sidesAxiom
ASA2 angles + included sideSide must be between the anglesTheorem 7.1
AAS2 angles + any one sideSide need not be includedTheorem 7.2
SSSAll 3 sides equalNoneTheorem 7.4
RHSHypotenuse + one leg + right angleOnly for right-angled trianglesTheorem 7.5

Invalid criteria — do NOT use:

  • SSA (two sides + non-included angle): ambiguous, two different triangles can satisfy the same SSA conditions.
  • AAA (all three angles equal): proves similarity only, not congruence (triangles can be scaled differently).

13. Common Mistakes to Avoid

  • Writing a congruence statement like △ABC ≅ △QPR without respecting vertex correspondence — always check that matching vertices are in matching positions.
  • Using SAS when the equal angle is not between the two named sides (that is SSA — not valid).
  • Applying RHS to a triangle that is not right-angled, or forgetting to identify which side is the hypotenuse.
  • Claiming AAA is a congruence criterion — it proves similarity, never congruence.
  • Deducing equal parts (CPCT) before writing the congruence statement. CPCT must come after the ≅ line.
  • In inequality proofs, mixing up Theorem 7.6 and its converse. State clearly: “greater side ⇒ greater opposite angle” vs “greater angle ⇒ greater opposite side.”
  • Forgetting that 1 + 2 > 3 (strict inequality) is required; equality gives a degenerate triangle (collinear points), not a real triangle.
Practice MCQs
1. In △ABC and △DEF, AB = DE, BC = EF and ∠B = ∠E. The congruence criterion used is:
  1. SSS
  2. ASA
  3. SAS
  4. RHS
Answer: (C) SAS — two sides AB = DE and BC = EF with included angle ∠B = ∠E between them.
2. In △ABC, AB = AC and ∠B = 55°. The measure of ∠A is:
  1. 55°
  2. 70°
  3. 110°
  4. 80°
Answer: (B) AB = AC ⇒ ∠B = ∠C = 55°. So ∠A = 180 − 55 − 55 = 70°.
3. CPCT stands for:
  1. Congruent Parts of Common Triangles
  2. Corresponding Parts of Congruent Triangles
  3. Common Parts of Congruent Terms
  4. Corresponding Pairs of Congruent Terms
Answer: (B) Corresponding Parts of Congruent Triangles are equal. This is the standard full form.
4. In △PQR, ∠P = 60°, ∠Q = 50°, ∠R = 70°. Which side is the longest?
  1. PQ
  2. QR
  3. PR
  4. All equal
Answer: (C) The largest angle is ∠R = 70°. The side opposite R is PQ... wait, in △PQR, side opposite ∠R is PQ. But let us recount: side PQ is opposite ∠R; side QR is opposite ∠P; side PR is opposite ∠Q. Largest angle ∠R = 70° ⇒ opposite side PQ is longest. So answer is (A) PQ.
5. Which of the following sets of lengths CAN form a triangle?
  1. 2 cm, 3 cm, 6 cm
  2. 1 cm, 2 cm, 3 cm
  3. 3 cm, 4 cm, 5 cm
  4. 4 cm, 5 cm, 10 cm
Answer: (C) Check 3 + 4 = 7 > 5 (yes). For (A): 2 + 3 = 5 < 6 (no). For (B): 1 + 2 = 3, not strictly greater (no). For (D): 4 + 5 = 9 < 10 (no). Only (C) satisfies the triangle inequality.
6. In a right-angled triangle the right angle is at C. Which inequality is always true?
  1. AB < BC
  2. AB < AC
  3. AB > AC and AB > BC
  4. AB = AC + BC
Answer: (C) AB is the hypotenuse (opposite the right angle, which is the largest angle). By Theorem 7.7, AB is greater than each of AC and BC.
7. If △ABC ≅ △FDE, which of the following is correct?
  1. AB = FD and BC = DE
  2. AB = FD and BC = FE
  3. AB = DF and BC = EF
  4. AB = EF and BC = DF
Answer: (A) Correspondence: A↔F, B↔D, C↔E. So AB = FD (vertices A and B match F and D) and BC = DE (vertices B and C match D and E).
8. The RHS congruence criterion can be applied only when:
  1. Both triangles are equilateral
  2. Both triangles are isosceles
  3. Both triangles are right-angled
  4. One angle in one triangle equals one angle in the other
Answer: (C) RHS requires a right angle (90°) in both triangles. “R” stands for the right angle present in each triangle.
9. In △ABC, AB = 5 cm, BC = 7 cm, CA = 6 cm. The greatest angle is:
  1. ∠A
  2. ∠B
  3. ∠C
  4. All equal
Answer: (A) The greatest side is BC = 7 cm. The angle opposite BC is ∠A (the angle at vertex A). So ∠A is greatest.
10. AAA (all three angles equal) proves that two triangles are:
  1. Congruent
  2. Similar but not necessarily congruent
  3. Identical in area
  4. Right-angled
Answer: (B) AAA proves similarity (same shape, same angles) but the triangles may have different sizes. Congruence also requires equal sides. So AAA is NOT a congruence criterion.
Assertion–Reason
A: If in △ABC, ∠B = ∠C, then AB = AC.   R: Sides opposite to equal angles of a triangle are equal.
Answer: Both A and R are true, and R is the correct explanation of A. This is Theorem 7.3 (NCERT): ∠B = ∠C ⇒ the sides opposite them (AC and AB respectively) are equal, so AB = AC.
A: SSA (two sides and a non-included angle) is a valid congruence criterion.   R: If two sides and one angle of one triangle equal two sides and one angle of another, the triangles must be congruent.
Answer: Both A and R are FALSE. SSA is an ambiguous case — the same two sides and non-included angle can produce two different (non-congruent) triangles. This is why NCERT does not include it as a criterion.
Previous-Year Questions
PYQ 1. Prove that angles opposite to equal sides of an isosceles triangle are equal. (CBSE, 3 marks, frequently asked)
Solution: Let △ABC with AB = AC. Construct bisector AD of ∠BAC meeting BC at D.
In △ABD and △ACD: AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common).
By SAS: △ABD ≅ △ACD. Therefore ∠ABD = ∠ACD by CPCT, i.e. ∠ABC = ∠ACB. ■
PYQ 2. In the figure, AC = BC, ∠DCA = ∠ECB and ∠DBC = ∠EAC. Prove DC = EC. (CBSE Board 2020, 3 marks)
Solution: ∠DCA = ∠ECB (given). Add ∠DCE to both sides:
∠DCA + ∠DCE = ∠ECB + ∠DCE ⇒ ∠ACE = ∠DCB.
In △ACE and △DCB: AC = BC (given), ∠ACE = ∠DCB (proved), ∠EAC = ∠DBC (given).
By AAS: △ACE ≅ △DCB. Therefore EC = BC... by CPCT: CE = CD, i.e. DC = EC. ■
PYQ 3. Show that in a right triangle, the hypotenuse is the longest side. (CBSE Board 2019, 2 marks)
Solution: Let △ABC with ∠B = 90°.
By angle sum: ∠A + ∠B + ∠C = 180° ⇒ ∠A + ∠C = 90°.
So ∠A < 90° = ∠B and ∠C < 90° = ∠B.
Thus ∠B is the largest angle. By Theorem 7.7, the side opposite ∠B (i.e. AC, the hypotenuse) is the longest. ■
PYQ 4. D is a point on side BC of △ABC such that AD = AC. Show AB > AD. (CBSE Board 2018, 3 marks)
Solution: In △ADC, AD = AC (given) ⇒ ∠ACD = ∠ADC (angles opposite equal sides).
∠ADB is an exterior angle of △ADC, so ∠ADB > ∠ACD (exterior angle > each non-adjacent interior angle).
Therefore ∠ADB > ∠ACD = ∠ADC ⇒ ∠ADB > ∠ABC (since ∠ABC = ∠ABD is the interior angle at B in △ABD).
In △ABD, ∠ADB > ∠ABD ⇒ AB > AD (side opposite greater angle is greater). ■
PYQ 5. In △ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC with DL = DM, prove AB = AC. (CBSE Board 2017, 3 marks)
Solution: In △DLB and △DMC:
∠DLB = ∠DMC = 90° (given perpendiculars), BD = CD (D is midpoint of BC), DL = DM (given).
By RHS: △DLB ≅ △DMC. Therefore ∠DBL = ∠DCM by CPCT, i.e. ∠ABC = ∠ACB.
By Theorem 7.3 (sides opposite equal angles are equal): AB = AC. ■
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