Quadrilaterals Exercise 8.2 Class 9 NCERT Solutions & Concepts

The Core Theorems: Mid-Point Theorem and its Converse

To solve any problem in class 9 maths quadrilaterals ex 8 2, you need to understand two foundational theorems:

1. 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. For a triangle $ABC$, if $D$ and $E$ are mid-points of $AB$ and $AC$ respectively, then $DE \parallel BC$ and $DE = \frac{1}{2} BC$.

1. Converse of the Mid-Point Theorem: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. In triangle $ABC$, if $D$ is the mid-point of $AB$ and a line $l$ is drawn through $D$ parallel to $BC$ intersecting $AC$ at $E$, then $AE = EC$ ($E$ is the mid-point of $AC$).

To apply these theorems to quadrilaterals, we draw a diagonal to split the quadrilateral into two triangles, allowing us to find pairs of parallel and equal lines.

The Step-by-Step Proof Blueprint

1. Identify the Triangles — Look for a diagonal (like AC) that splits the quadrilateral into two distinct triangles (like ADC and ABC).
2. Apply Mid-Point Theorem to Triangle 1 — In triangle ADC, locate the mid-points S and R of sides AD and CD. By Mid-point theorem, SR || AC and SR = 1/2 AC.
3. Apply Mid-Point Theorem to Triangle 2 — In triangle ABC, locate the mid-points P and Q of sides AB and BC. By Mid-point theorem, PQ || AC and PQ = 1/2 AC.
4. Compare and Conclude — Since SR || AC and PQ || AC, it implies SR || PQ. Similarly, since SR = 1/2 AC and PQ = 1/2 AC, we get SR = PQ. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram!

Standard Proof: Midpoints of a Quadrilateral

• Example: Show that the quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a parallelogram.
• Step 1: Let ABCD be a quadrilateral. P, Q, R, and S are the mid-points of AB, BC, CD, and DA. Join AC (the diagonal).
• Step 2: In Triangle ADC: S and R are midpoints of AD and CD. Therefore, by the Mid-Point Theorem, SR || AC and SR = 1/2 AC (Equation 1).
• Step 3: In Triangle ABC: P and Q are midpoints of AB and BC. Therefore, by the Mid-Point Theorem, PQ || AC and PQ = 1/2 AC (Equation 2).
• Step 4: From Equations 1 and 2, we get SR || PQ and SR = PQ.
• Step 5: Since one pair of opposite sides (PQ and SR) of quadrilateral PQRS is equal and parallel, PQRS is a parallelogram. (Hence proved)

Common Mistakes & CBSE Board Tips

• Forget to Construct Diagonals: Most students get stuck because they try to prove relationships directly between the midpoints. Remember: Always draw a diagonal (or both diagonals) first!
• Incomplete Statements: Writing 'by mid-point theorem' without stating the specific triangle you are applying it to will cost you marks. State clearly: 'In Triangle ABC, by Mid-Point Theorem...'.
• Confusing Rectangle and Rhombus Properties: Remember that if the diagonals of the outer quadrilateral are perpendicular (like in a rhombus), the inner midpoint quadrilateral is a rectangle. If the outer diagonals are equal (like in a rectangle), the inner midpoint quadrilateral is a rhombus.

Practice Questions with Solutions

• Q: ABCD is a rectangle and P, Q, R, S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. A: Step 1: Draw the rectangle ABCD and its diagonals AC and BD. Since ABCD is a rectangle, its diagonals are equal: AC = BD. Step 2: In Triangle ADC, S and R are midpoints of AD and CD. Thus, SR || AC and SR = 1/2 AC. Step 3: In Triangle ABC, P and Q are midpoints of AB and BC. Thus, PQ || AC and PQ = 1/2 AC. Therefore, SR || PQ and SR = PQ. Step 4: Similarly, looking at diagonal BD, in Triangle BCD, QR || BD and QR = 1/2 BD. Step 5: Since AC = BD, we have 1/2 AC = 1/2 BD, which means SR = PQ = QR = SP. Final answer: Since all sides of parallelogram PQRS are equal, PQRS is a rhombus.
• Q: ABCD is a rhombus and P, Q, R, S are the midpoints of sides AB, BC, CD, DA. Show that PQRS is a rectangle. A: Step 1: Draw rhombus ABCD and its diagonals AC and BD. Diagonals of a rhombus intersect at 90 degrees. Step 2: Since P, Q, R, S are midpoints, we have PQ || AC (from Triangle ABC) and SP || BD (from Triangle ABD). Step 3: This means the sides of PQRS are parallel to the diagonals of ABCD. Step 4: Since the diagonals AC and BD intersect at 90 degrees, the parallel sides PQ and SP must also meet at 90 degrees. Final answer: Since PQRS is a parallelogram with one angle equal to 90 degrees, PQRS is a rectangle.
• Q: 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. A: Step 1: Let the line EF intersect the diagonal BD at point G. Step 2: In Triangle ABD, E is the midpoint of AD and EG || AB (since EF || AB). Step 3: By the Converse of the Mid-Point Theorem, G must be the midpoint of diagonal BD. Step 4: Now, look at Triangle BCD. We know G is the midpoint of BD and GF || AB || DC. Step 5: Therefore, in Triangle BCD, a line GF is drawn through the midpoint G parallel to DC. Final answer: By the Converse of the Mid-Point Theorem, F is the mid-point of BC.
• Q: Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other. A: Step 1: Let ABCD be a quadrilateral. P, Q, R, S are midpoints of AB, BC, CD, DA. Step 2: The line segments joining opposite midpoints are PR and QS. Step 3: Join PQ, QR, RS, and SP to form a quadrilateral PQRS. Step 4: As proven in standard Mid-Point proofs, PQRS is a parallelogram because PQ || SR and PQ = SR. Step 5: We know that the diagonals of a parallelogram bisect each other. Final answer: Since PR and QS are the diagonals of the parallelogram PQRS, they must bisect each other.

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