Important Formulas, Rules & Key Points

Until now, we have studied the properties of equality in triangles (like congruency). Now, let us understand the relations of inequality between angles and sides of a triangle:

• Theorem 7.6 (Angle-Side Inequality 1): If two sides of a triangle are unequal, the angle opposite to the longer side is larger (greater).
• Theorem 7.7 (Angle-Side Inequality 2): In any triangle, the side opposite to the larger (greater) angle is longer.
• Theorem 7.8 (Triangle Inequality Theorem): The sum of any two sides of a triangle is strictly greater than the third side.
  Example: For any triangle ABC, AB + BC > AC, BC + AC > AB, and AC + AB > BC.
• Shortest Distance Rule: Of all line segments drawn from a given point to a given line, the perpendicular line segment is the shortest.

Question 1

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

Proof:

1. Let ΔABC be a right-angled triangle, right-angled at B (i.e., ∠B = 90°).
2. By the Angle Sum Property of a triangle:
   ∠A + ∠B + ∠C = 180°
3. Substitute ∠B = 90° into the equation:
   ∠A + 90° + ∠C = 180°
∠A + ∠C = 90°
4. Since the sum of ∠A and ∠C is 90°, both angles must be acute (less than 90°).
   Therefore, ∠B > ∠A and ∠B > ∠C.
5. We know that the side opposite to the larger angle is longer. Thus:
   • Since ∠B > ∠A ⇒ AC > BC
   • Since ∠B > ∠C ⇒ AC > AB
6. Since AC (the hypotenuse) is greater than both AB and BC, the hypotenuse is indeed the longest side of a right-angled triangle.
   Hence Proved!

Question 2

In the given figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Proof:

1. Observe that ∠ABC and ∠PBC form a linear pair on line AP.
   Therefore, ∠ABC + ∠PBC = 180° ⇒ ∠ABC = 180° - ∠PBC — (Equation 1)
2. Similarly, ∠ACB and ∠QCB form a linear pair on line AQ.
   Therefore, ∠ACB + ∠QCB = 180° ⇒ ∠ACB = 180° - ∠QCB — (Equation 2)
3. We are given that:
   ∠PBC < ∠QCB
4. Multiplying both sides of the inequality by -1 reverses the inequality sign:
   -∠PBC > -∠QCB
5. Adding 180° to both sides:
   180° - ∠PBC > 180° - ∠QCB
6. Using Equation 1 and Equation 2, we can rewrite this as:
   ∠ABC > ∠ACB
7. Since the side opposite to the larger angle is longer:
   AC > AB
   Hence Proved!

Question 3

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Proof:

1. Let O be the point of intersection of lines AD and BC.
2. Consider ΔOAB:
   Given that ∠B < ∠A (or ∠A > ∠B).
   Since the side opposite to the smaller angle is shorter:
   OA < OB — (Equation 1)
3. Consider ΔOCD:
   Given that ∠C < ∠D (or ∠D > ∠C).
   Therefore:
   OD < OC — (Equation 2)
4. Adding Equation 1 and Equation 2:
   OA + OD < OB + OC
5. From the figure, we see that OA + OD = AD and OB + OC = BC. Therefore:
   AD < BC
   Hence Proved!

Question 4

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Proof:

1. To prove ∠A > ∠C:
   Join diagonal AC. Let us label the angles: ∠BAC = ∠1, ∠CAD = ∠3, ∠BCA = ∠2, and ∠ACD = ∠4.
2. In ΔABC:
   Since AB is the smallest side of the quadrilateral, BC > AB.
   The angle opposite to the longer side is larger, so:
   ∠1 > ∠2 — (Equation 1)
3. In ΔADC:
   Since CD is the longest side of the quadrilateral, CD > AD.
   Therefore:
   ∠3 > ∠4 — (Equation 2)
4. Adding Equation 1 and Equation 2:
   ∠1 + ∠3 > ∠2 + ∠4
   Since ∠1 + ∠3 = ∠A and ∠2 + ∠4 = ∠C, we get:
   ∠A > ∠C
5. To prove ∠B > ∠D:
   Similarly, join diagonal BD. By applying the same logic to ΔABD and ΔBCD, we get:
   ∠B > ∠D.
   Hence Proved!

Question 5

In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Proof:

1. In ΔPQR, we are given:
   PR > PQ
   Therefore, the angle opposite to side PR is greater than the angle opposite to PQ:
   ∠PQR > ∠PRQ — (Equation 1)
2. Since PS is the angle bisector of ∠QPR:
   ∠QPS = ∠RPS — (Equation 2)
3. Now, let's use the exterior angle property:
   • For ΔPQS, the exterior angle is ∠PSR:
     ∠PSR = ∠PQR + ∠QPS — (Equation 3)
   • For ΔPRS, the exterior angle is ∠PSQ:
     ∠PSQ = ∠PRQ + ∠RPS — (Equation 4)
4. From Equation 1, we know that ∠PQR > ∠PRQ. Adding the equal terms ∠QPS and ∠RPS (from Equation 2) to both sides of this inequality:
   ∠PQR + ∠QPS > ∠PRQ + ∠RPS
5. Substituting from Equation 3 and Equation 4:
   ∠PSR > ∠PSQ
   Hence Proved!

Question 6

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Proof:

1. Let l be a straight line and P be a point not lying on line l.
2. Draw a perpendicular PM from point P to line l, meeting l at M (i.e., ∠M = 90°).
3. Let N be any other point on line l other than M. Join PN to form ΔPMN.
4. In right-angled triangle PMN:
   ∠M = 90°
   By the angle sum property, the remaining two angles must be acute:
   ∠N < 90°
5. Therefore, ∠M > ∠N.
6. Since the side opposite to the larger angle is longer:
   PN > PM (or PM < PN).
7. Since N can be any arbitrary point on line l, PM is shorter than any other line segment connecting P to line l.
   Thus, the perpendicular segment is the shortest.
   Hence Proved!

Aptitude Corner (For Olympiads & NTSE)

Triangle inequalities are a favorite topic in competitive examinations. Keep these shortcuts in mind:

• Range of the third side: If two sides of a triangle are a and b (where a ≥ b), then the third side c must satisfy:
  (a - b) < c < (a + b)
• Quick Quiz Question: If two sides of an isosceles triangle are 4 cm and 9 cm, what is the length of the third side?
  Solution: The sides must be either (4, 4, 9) or (4, 9, 9). But for (4, 4, 9), 4 + 4 = 8 which is less than 9 (violating the triangle inequality theorem). Thus, the third side must be 9 cm.