Important Formulas, Rules & Key Points

• The Mid-point Theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
  In ΔABC, if D and E are mid-points of AB and AC, then DE || BC and DE = ½ BC.
• Converse of Mid-point Theorem: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
• Parallelogram Property: A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.
• Diagonals of a Rhombus: The diagonals of a rhombus bisect each other at right angles (90°).
• Diagonals of a Rectangle: The diagonals of a rectangle are equal in length and bisect each other.

Question 1: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:

(i) SR || AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Proof:

(i) In ΔADC,
S is the mid-point of AD and R is the mid-point of CD (Given).
By the Mid-point Theorem, the line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
Therefore, SR || AC and SR = ½ AC — (Equation 1)

(ii) In ΔABC,
P is the mid-point of AB and Q is the mid-point of BC (Given).
By the Mid-point Theorem:
PQ || AC and PQ = ½ AC — (Equation 2)
From Equation 1 and Equation 2, we get:
PQ = SR (Since both are equal to ½ AC).

(iii) From Equation 1 and Equation 2, we also have:
PQ || AC and SR || AC ⇒ PQ || SR.
We already proved that PQ = SR.
Since one pair of opposite sides of quadrilateral PQRS is equal and parallel, PQRS is a parallelogram.
Hence Proved.

Question 2: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Proof:

First, we join the diagonals AC and BD of the rhombus ABCD. Let them intersect at point O.

1. In ΔADC, S and R are mid-points of AD and CD. By Mid-point Theorem:
   SR || AC and SR = ½ AC — (Equation 1)
2. In ΔABC, P and Q are mid-points of AB and BC. By Mid-point Theorem:
   PQ || AC and PQ = ½ AC — (Equation 2)
3. From (1) and (2), we get PQ || SR and PQ = SR.
   Thus, PQRS is a parallelogram.
4. Now, we need to prove that one angle of this parallelogram is 90°.
   Let the line segment PQ intersect BD at M and SR intersect BD at N (or let the small intersection points on the diagonals be called E and F). Since ABCD is a rhombus, the diagonals intersect at 90° ⇒ ∠AOD = 90°.
   Since SP || BD (by Mid-point theorem in ΔABD) and PQ || AC:
   The smaller quadrilateral formed by the intersecting lines is also a parallelogram, which means its opposite angles are equal.
   Therefore, ∠SPQ = ∠AOD = 90°.

Since PQRS is a parallelogram with one angle equal to 90°, PQRS is a rectangle.
Hence Proved.

Question 3: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Proof:

To show PQRS is a rhombus, we need to show it is a parallelogram and all its sides are equal.

1. By applying the Mid-point Theorem in ΔABC and ΔADC, we get:
   PQ || AC, PQ = ½ AC and SR || AC, SR = ½ AC.
   Therefore, PQ || SR and PQ = SR. This makes PQRS a parallelogram.
2. Similarly, by joining diagonal BD and applying Mid-point Theorem in ΔABD and ΔBCD:
   PS = ½ BD and QR = ½ BD.
3. Since ABCD is a rectangle, its diagonals are equal: AC = BD.
   ⇒ ½ AC = ½ BD
   ⇒ PQ = PS.
4. In parallelogram PQRS, adjacent sides PQ and PS are equal. Thus, all four sides PQ, QR, RS, and SP are equal.
   Hence, PQRS is a rhombus.
   Hence Proved.

Question 4: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

Proof:

Let the line EF intersect the diagonal BD at point G.

1. In ΔDAB:
   E is the mid-point of AD (Given).
   EG || AB (Since EF || AB and G lies on EF).
   By the Converse of the Mid-point Theorem, a line drawn through the mid-point of one side parallel to another side bisects the third side.
   Therefore, G is the mid-point of BD.
2. In ΔBCD:
   G is the mid-point of BD (Proved above).
   GF || DC (Since AB || DC and EF || AB, which means EF || DC).
   By the Converse of the Mid-point Theorem, the line through G parallel to DC must bisect BC.
   Therefore, F is the mid-point of BC.
   Hence Proved.

Question 5: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

Proof:

To prove that AF and EC trisect BD, we need to show that DP = PQ = QB, where P and Q are the points of intersection of BD with AF and EC respectively.

1. Since ABCD is a parallelogram, AB || CD and AB = CD.
   ⇒ AE || FC (parts of parallel lines) and ½ AB = ½ CD ⇒ AE = FC.
   Since AE || FC and AE = FC, AECF is a parallelogram.
   Therefore, AF || EC. This also means AP || EQ and FP || CQ.
2. In ΔDQC:
   F is the mid-point of DC (Given).
   FP || CQ (Since AF || EC).
   By the Converse of Mid-point Theorem, P is the mid-point of DQ.
   ⇒ DP = PQ — (Equation 1)
3. In ΔABP:
   E is the mid-point of AB (Given).
   EQ || AP (Since AF || EC).
   By the Converse of Mid-point Theorem, Q is the mid-point of BP.
   ⇒ PQ = QB — (Equation 2)
4. From Equation 1 and Equation 2:
   DP = PQ = QB.
   Thus, the line segments AF and EC trisect the diagonal BD.
   Hence Proved.

Question 6: Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Proof:

1. Let ABCD be a quadrilateral. P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively.
2. We join PR and SQ (the lines joining opposite mid-points) and also join PQ, QR, RS, and SP.
3. In ΔABC, P and Q are mid-points ⇒ PQ || AC and PQ = ½ AC (Mid-point Theorem).
4. In ΔADC, S and R are mid-points ⇒ SR || AC and SR = ½ AC (Mid-point Theorem).
5. Therefore, PQ || SR and PQ = SR. Hence, PQRS is a parallelogram.
6. The diagonals of a parallelogram bisect each other.
   Here, the diagonals of the parallelogram PQRS are PR and SQ.
   Therefore, PR and SQ bisect each other.
   Hence Proved.

Question 7: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that:

(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB

Proof:

(i) In ΔACB:
M is the mid-point of AB (Given).
MD || BC (Given).
By the Converse of the Mid-point Theorem, D must be the mid-point of AC.
Hence, D is the mid-point of AC.

(ii) Since MD || BC:
∠ADM = ∠ACB (Corresponding angles).
Since ∠ACB = 90° (Given), we have:
∠ADM = 90°.
Therefore, MD ⊥ AC.

(iii) In ΔAMD and ΔCMD:
AD = CD (Since D is the mid-point of AC).
∠ADM = ∠CDM = 90° (Since MD ⊥ AC).
MD = MD (Common side).
Therefore, by SAS congruence criterion, ΔAMD ≅ ΔCMD.
By CPCT (Corresponding Parts of Congruent Triangles):
CM = MA.
Since M is the mid-point of AB, we have MA = ½ AB.
Therefore, CM = MA = ½ AB.
Hence Proved.

💡 Aptitude Corner (For NTSE & Olympiad Exams)

Varignon's Theorem: The quadrilateral formed by connecting the midpoints of consecutive sides of any quadrilateral is always a parallelogram. Its area is always exactly half of the area of the outer quadrilateral.

Quick Formula: Area(PQRS) = ½ × Area(ABCD)