TOPICS

Exercise - 8.2

Quadrilaterals

**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 Figure). AC is a diagonal. Show that :

(i) SR || AC and SR = AC/2

(ii) PQ = SR

(iii) PQRS is a parallelogram

(i) Given that: In ∆ADC, S and R are the mid-points of sides AD and CD respectively. Proove that : SR || AC and SR = AC/2 Proof : In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. So, SR || AC and SR = AC/2. .......(1) (ii) Given that: In ∆ABC, P and Q are mid-points of sides AB and BC respectively. Proove that : PQ = SR Proof : Therefore, by using mid-point theorem, PQ || AC and PQ = AC ... (2) Using equations (1) and (2), we obtain PQ || SR and PQ = SR ... (3) So, PQ = SR (iii) Proove that : PQRS is a parallelogram Proof : From equation (3), we obtained PQ || SR and PQ = SR Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram.

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

Given that : In ∆ABC, P and Q are the mid-points of sides AB and BC respectively. Prove that : Quadrilateral PQRS is a rectangle. Proof : In ∆ABC, R and S are the mid-points of CD and AD respectively. PQ || AC and PQ = AC/2 (Using mid-point theorem......(1) RS || AC and RS = AC/2 (Using mid-point theorem).....(2) In ∆ADC, From equations (1) and (2), we obtain PQ || RS and PQ = RS Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. In quadrilateral OMQN, MQ || ON ( PQ || AC) QN || OM ( QR || BD) Therefore, OMQN is a parallelogram. ∠MQN = ∠NOM ∠PQR = ∠NOM However, NOM = 90° (Diagonals of a rhombus are perpendicular to each other) ∠PQR = 90° Clearly, PQRS is a parallelogram having one of its interior angles as 90°. Hence, PQRS is a rectangle.

**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.

Given that : In ∆ABC, P and Q are the mid-points of AB and BC respectively. Prove that : ABCD is a rhombus or ABCD is a parallelogram and all the sides of ABCD are equal. Construction : Let us join AC and BD. Proof : In ∆ABC, PQ || AC and PQ = AC/2 (Mid-Point Theorem).....(1) Similarly in ∆ADC, SR || AC and SR = AC (Mid-point theorem) ... (2) Clearly, PQ || SR and PQ = SR Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, It is a parallelogram. PS || QR and PS = QR (Opposite sides of parallelogram)... (3) In ∆BCD, Q and R are the mid-points of side BC and CD respectively. QR || BD and QR = BD (Mid-point theorem) ... (4) However, the diagonals of a rectangle are equal. AC = BD .........(5) By using equation (1), (2), (3), (4), and (5), we obtain PQ = QR = SR = PS Therefore, PQRS is a rhombus.

**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 (see Figure). Show that F is the mid-point of BC.

Given that : Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. Prove that : F is the mid-point of BC. Proof : In ∆ABD, EF || AB and E is the mid-point of AD. Therefore, G will be the mid-point of DB. As EF || AB and AB || CD, EF || CD (Two lines parallel to the same line are parallel to each other) In ∆BCD, GF || CD and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC.

**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.

Given that : ABCD is a parallelogram. AB || CD And hence, AE || FC Again, AB = CD (Opposite sides of parallelogram ABCD) AB/2 = CD/2 AE = FC (E and F are mid-points of side AB and CD) Prove that : The line segments AF and EC trisect the diagonal BD. Proof : In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. AF || EC (Opposite sides of a parallelogram) In ∆DQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ. DP = PQ ... (1) Similarly, in ∆APB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB. PQ = QB ... (2) From equations (1) and (2), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD.

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

Given that : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Prove that : PR and QS bisect each other. Construction : Join PQ, QR, RS, SP, and BD. Proof : In ∆ABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that SP || BD and SP = BD/2 ... (1) Similarly in ∆BCD, QR || BD and QR = BD/2 ... (2) From equations (1) and (2), we obtain SP || QR and SP = QR In quadrilateral SPQR, one pair of opposite sides is equal and parallel to each other. Therefore, SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other.

**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 = AB/2

Given that : ABCD is a rhombus. Prove that : (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = AB/2 Proof : (i) In ∆ABC, It is given that M is the mid-point of AB and MD || BC. Therefore, D is the mid-point of AC. (Converse of mid-point theorem) (ii) As DM || CB and AC is a transversal line for them, therefore, ∠MDC + ∠DCB = 180° (Co-interior angles) ∠MDC + 90° = 180° ∠MDC = 90° MD ⊥ AC (iii) Join MC. In ∆AMD and ∆CMD, AD = CD (D is the mid-point of side AC) ∠ADM = ∠CDM (Each 90°) DM = DM (Common) ∆AMD ≅ ∆CMD (By SAS congruence rule) Therefore, AM = CM (By CPCT) However, AM = AB (M is the mid-point of AB) Therefore, it can be said that CM = AM = AB/2

CLASSES