GSEB Class 9 Maths Solutions Chapter 8 Quadrilaterals Exercise 8.2

 

Question 1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (See fig.) AC is a diagonal. Show that 1. SR || AC and SR = \( \frac {1}{2} \) AC 2. PQ = SR 3. PQRS is a parallelogram.

Answer:1. In triangle ADC, S and R are the middle points of sides AD and DC respectively.
\( \implies \) SR is parallel to AC and SR = \( \frac {1}{2} \) AC .......(1) (A line segment joining the middle points of two sides of a triangle is parallel to the third side and is half of its length.) 2. In triangle ABC, P and Q are the middle points of sides AB and BC respectively.
\( \implies \) PQ is parallel to AC and PQ = \( \frac {1}{2} \) AC ......(2) (A line segment joining the middle points of two sides of a triangle is parallel to the third side and is half of its length.) From formulas (1) and (2), we get PQ = SR. 3. From formulas (1) and (2), we have PQ is parallel to SR and PQ = SR. Therefore, PQRS is a parallelogram. (If one set of facing sides of a quadrilateral are equal in length and run side-by-side, then it is a parallelogram).
In simple words: We use the midpoint theorem twice, once for SR and once for PQ, relating them to AC. This shows SR and PQ are both half of AC and parallel to AC, meaning they are equal and parallel to each other. This condition then proves that PQRS is a parallelogram.

Exam Tip: Remember to clearly state the midpoint theorem when using it in your proof. Label your equations for easy referencing later.

 

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.

Answer:Given: ABCD is a rhombus, and P, Q, R, S are the middle points of AB, BC, CD, DA in order. PQ, QR, RS, and SP are joined. To Prove: PQRS is a rectangle. Construction: Join AC and BD. Proof: In triangle ADC, S and R are the middle points of AD and DC.
\( \implies \) SR is parallel to AC and SR = \( \frac {1}{2} \) AC .......(1) (A line segment joining the middle points of two sides of a triangle is parallel to the third side and is half of its length.) Similarly, in triangle ABC, P and Q are the middle points of AB and BC.
\( \implies \) PQ is parallel to AC and PQ = \( \frac {1}{2} \) AC ...(2) From formulas (1) and (2), we get PQ = SR and PQ is parallel to SR. Therefore, PQRS is a parallelogram. Now, the diagonals of a rhombus cross each other at 90°. Let the diagonals AC and BD intersect at O. Consider the figure where SP and PQ are sides of PQRS. Let SP intersect BD at F and PQ intersect AC at E. Since SP || BD and PQ || AC, and AC \( \perp \) BD. We have SP || FO and PQ || EO. Also, in parallelogram OESF, if \( \angle \) EOF = 90°, then \( \angle \) S = 90°.
\( \implies \) \( \angle \) DFS = \( \angle \) EOF = 90° (These are corresponding angles). Again, SP is parallel to BD. Therefore, \( \angle \) ESF = \( \angle \) DFS (These are alternate interior angles).
\( \implies \) \( \angle \) ESF = 90°. Hence, PQRS is a rectangle.
In simple words: First, we show PQRS is a parallelogram using the midpoint theorem. Then, because ABCD is a rhombus, its diagonals (AC and BD) cross at a 90-degree angle. By using properties of parallel lines and corresponding/alternate angles, we prove that one angle of the parallelogram PQRS is 90 degrees. A parallelogram with a 90-degree angle is a rectangle.

Exam Tip: Remember that the diagonals of a rhombus bisect each other at right angles. This property is key to showing that the interior angles of PQRS are 90 degrees.

 

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

Answer:Given: ABCD is a rectangle. P, Q, R, and S are the middle points of AB, BC, CD, and DA respectively. To Prove: Quadrilateral PQRS is a rhombus. Construction: Join AC and BD. Proof: In triangle ADC, S and R are the middle points of AD and DC.
\( \implies \) SR is parallel to AC and SR = \( \frac {1}{2} \) AC .......(1) (A line segment joining the middle points of two sides of a triangle is parallel to the third side and is half of its length.) Similarly, in triangle ABC, P and Q are the middle points of AB and BC.
\( \implies \) PQ is parallel to AC and PQ = \( \frac {1}{2} \) AC .......(2) From formulas (1) and (2), we get PQ is parallel to SR and PQ = SR. Thus, PQRS is a parallelogram. Now, we know that the diagonals of a rectangle are equal in length. Therefore, AC = BD.
\( \implies \) \( \frac {1}{2} \) AC = \( \frac {1}{2} \) BD. From formula (2), PQ = \( \frac {1}{2} \) AC. Also, in triangle BCD, Q and R are the middle points of BC and CD.
\( \implies \) QR = \( \frac {1}{2} \) BD. Since \( \frac {1}{2} \) AC = \( \frac {1}{2} \) BD, it follows that PQ = QR. A parallelogram with adjacent sides of equal length is a rhombus. Therefore, PQRS is a rhombus.
In simple words: First, we use the midpoint theorem to show that PQRS is a parallelogram. Since ABCD is a rectangle, its diagonals (AC and BD) are equal in length. Using the midpoint theorem again for adjacent sides of PQRS (like PQ and QR) and relating them to the equal diagonals, we prove that adjacent sides of PQRS are equal. A parallelogram with equal adjacent sides is a rhombus.

Exam Tip: To prove a quadrilateral is a rhombus, first prove it's a parallelogram, then show that any pair of adjacent sides are equal. The property of a rectangle having equal diagonals is crucial here.

 

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

Answer:To Prove: F is the middle point of BC. Proof: Let DB intersect EF at G. In triangle DAB, E is the middle point of DA, and EG is parallel to AB.
\( \implies \) G is the middle point of DB. (A line drawn from the middle point of one side and parallel to another side, divides the third side into two equal parts.) Now in triangle DBC, G is the middle point of DB, and GF is parallel to AB, which is also parallel to DC (since AB || DC and EF || AB).
\( \implies \) F is the middle point of BC. (A line drawn from the middle point of one side and parallel to another side divides the third side into two equal parts.)
In simple words: We use the converse of the midpoint theorem in two steps. First, in triangle DAB, since E is the midpoint of DA and EG is parallel to AB, G must be the midpoint of DB. Then, in triangle DBC, since G is the midpoint of DB and GF is parallel to DC, F must be the midpoint of BC.

Exam Tip: When using the converse of the midpoint theorem, ensure you explicitly state that the line passes through a midpoint and is parallel to another side, as this is the condition for it to bisect the third side.

 

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

Answer:To Prove: DP = PQ = BQ. Proof: In parallelogram ABCD, we know that facing sides are parallel and equal. So, AB is parallel to DC and AB = DC.
\( \implies \) \( \frac {1}{2} \) AB is parallel to \( \frac {1}{2} \) DC and \( \frac {1}{2} \) AB = \( \frac {1}{2} \) DC (by dividing both sides by 2). Since E and F are the middle points of AB and CD respectively, AE = \( \frac {1}{2} \) AB and FC = \( \frac {1}{2} \) DC. Therefore, AE is parallel to FC and AE = FC. Hence, AECF is a parallelogram. (If one pair of facing sides of a quadrilateral are equal and run side-by-side, then it is a parallelogram.) Since AECF is a parallelogram, its facing sides AF and EC are parallel.
\( \implies \) AF is parallel to CE. Now consider triangle DQC. F is the middle point of DC, and FP is parallel to QC (since AF || EC).
\( \implies \) DP = PQ .......(1) (A line drawn from the middle point of one side and parallel to the other side divides the third side into two equal parts.) Next, consider triangle ABP. E is the middle point of AB, and EQ is parallel to AP (since AF || EC).
\( \implies \) BQ = PQ .......(2) (A line drawn from the middle point of one side and parallel to the other side divides the third side into two equal parts.) From formulas (1) and (2), we can conclude that DP = PQ = BQ. Therefore, the diagonal BD is divided into three equal segments (trisected) by the line segments AF and CE.
In simple words: First, we show that AECF is a parallelogram because AE is parallel and equal to FC. This means AF is parallel to EC. Then, by applying the converse of the midpoint theorem in triangle DQC, we prove that DP = PQ. Similarly, in triangle ABP, we prove that BQ = PQ. Combining these, we find that DP, PQ, and BQ are all equal, meaning the diagonal BD is trisected.

Exam Tip: The key to this proof is recognizing that AECF is a parallelogram, which establishes the parallel lines AF and EC. Then, repeatedly apply the converse of the midpoint theorem to the two triangles formed on either side of the diagonal BD.

 

Question 6. Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

Answer:Given: A quadrilateral ABCD, in which P, Q, R, and S are the middle points of the sides AB, BC, DC, and AD respectively. To Prove: PR and QS divide each other into two equal parts. Construction: Join PQ, QR, RS, SP, AC, and BD. Proof: In triangle ABD, P and S are the middle points of AB and AD respectively.
\( \implies \) PS is parallel to BD and PS = \( \frac {1}{2} \) BD ......(1) (A line segment joining the middle points of two sides of a triangle is parallel to the third side and is half of its length.) Similarly, in triangle BCD, Q and R are the middle points of BC and CD respectively.
\( \implies \) QR is parallel to BD and QR = \( \frac {1}{2} \) BD ......(2) From formulas (1) and (2), we have PS is parallel to QR and PS = QR. If one set of facing sides of a quadrilateral are equal in length and run side-by-side, then it is a parallelogram. Therefore, PQRS is a parallelogram. Since PR and QS are the diagonals of the parallelogram PQRS, they must divide each other into two equal parts. (Diagonals of a parallelogram divide each other into two equal parts).
In simple words: We first prove that the quadrilateral formed by joining the midpoints of the sides of any quadrilateral (PQRS) is a parallelogram. We do this by applying the midpoint theorem to show that PS and QR are both parallel to and half the length of diagonal BD, making PS parallel and equal to QR. Once PQRS is shown to be a parallelogram, its diagonals (PR and QS) are known to bisect each other, which means they cut each other exactly in half.

Exam Tip: This is a standard proof using the midpoint theorem. The crucial step is establishing PQRS as a parallelogram. Make sure to clearly state both parts of the midpoint theorem (parallelism and half the length) and the property of parallelogram diagonals.

 

Question 7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D. Show that (i) D is the mid-point of AC. (ii) MD\( \perp \)AC. (iii) CM = MA = \( \frac {1}{2} \) AB.

Answer:(i) In triangle ABC, M is the middle point of AB, and MD is parallel to BC.
\( \implies \) D is the middle point of AC. (A line drawn through the middle point of one side and parallel to another side divides the third side into two equal parts.) (ii) Since MD is parallel to BC, and AC is a transversal line. The sum of consecutive interior angles is 180°.
\( \implies \) \( \angle \) MDC + \( \angle \) BCD = 180°. Given that \( \angle \) BCD = 90° (since the triangle is right-angled at C). So, \( \angle \) MDC + 90° = 180°.
\( \implies \) \( \angle \) MDC = 180° - 90° = 90°. Therefore, MD is perpendicular to AC. (iii) In triangle ADM and triangle CDM, AD = CD (This was proven in part (i)). \( \angle \) ADM = \( \angle \) CDM = 90° (Both are right angles because MD \( \perp \) AC, proven in part (ii)). DM = DM (This is a common side to both triangles). Hence, triangle ADM is congruent to triangle CDM (by the SAS (Side-Angle-Side) congruency rule).
\( \implies \) MA = CM (by CPCTC - Corresponding Parts of Congruent Triangles are Congruent). Also, M is the middle point of AB. Therefore, MA = \( \frac {1}{2} \) AB. Combining these, we get CM = MA = \( \frac {1}{2} \) AB.
In simple words: For (i), we use the converse of the midpoint theorem: since M is the midpoint of AB and MD is parallel to BC, D must be the midpoint of AC. For (ii), because MD is parallel to BC and AC is a straight line cutting them, and angle C is 90 degrees, angle MDC must also be 90 degrees, meaning MD is perpendicular to AC. For (iii), by proving triangles ADM and CDM are identical (congruent) using SAS, we show CM = MA. Since M is the midpoint of AB, MA is half of AB, so CM is also half of AB.

Exam Tip: This question combines several geometric concepts. Break down the proof into sub-parts, clearly stating the theorem or property used for each step. For part (iii), remember that the median to the hypotenuse in a right-angled triangle is half the length of the hypotenuse.