NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 3

Converse of mid-point theorem in triangel PQR

Converse of Mid-Point Theorem

Converse of mid-point theorem in triangel PQR

Converse of mid-point theorem: In a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its at mid-point.

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

Quadrilateral poly1 Quadrilateral poly1: Polygon D, F, E, A Quadrilateral poly2 Quadrilateral poly2: Polygon F, C, B, E Segment a Segment a: Segment [D, F] of Quadrilateral poly1 Segment a Segment a: Segment [D, F] of Quadrilateral poly1 Segment a Segment a: Segment [D, F] of Quadrilateral poly1 Segment b Segment b: Segment [F, E] of Quadrilateral poly1 Segment c Segment c: Segment [E, A] of Quadrilateral poly1 Segment c Segment c: Segment [E, A] of Quadrilateral poly1 Segment d Segment d: Segment [A, D] of Quadrilateral poly1 Segment b_1 Segment b_1: Segment [F, C] of Quadrilateral poly2 Segment b_1 Segment b_1: Segment [F, C] of Quadrilateral poly2 Segment b_1 Segment b_1: Segment [F, C] of Quadrilateral poly2 Segment e Segment e: Segment [C, B] of Quadrilateral poly2 Segment f Segment f: Segment [B, E] of Quadrilateral poly2 Segment f Segment f: Segment [B, E] of Quadrilateral poly2 Segment c_1 Segment c_1: Segment [E, F] of Quadrilateral poly2 Segment g Segment g: Segment [A, F] Segment h Segment h: Segment [E, C] Segment i Segment i: Segment [D, B] Point D D = (-2.42, 3.36) Point D D = (-2.42, 3.36) Point D D = (-2.42, 3.36) Point F F = (-0.04, 3.36) Point F F = (-0.04, 3.36) Point F F = (-0.04, 3.36) Point E E = (-0.88, 0.78) Point E E = (-0.88, 0.78) Point E E = (-0.88, 0.78) Point A A = (-3.32, 0.8) Point A A = (-3.32, 0.8) Point A A = (-3.32, 0.8) Point C C = (2.32, 3.36) Point C C = (2.32, 3.36) Point C C = (2.32, 3.36) Point B B = (1.5, 0.76) Point B B = (1.5, 0.76) Point B B = (1.5, 0.76) Point P Point P: Intersection point of g, i Point P Point P: Intersection point of g, i Point P Point P: Intersection point of g, i Point Q Point Q: Intersection point of h, i Point Q Point Q: Intersection point of h, i Point Q Point Q: Intersection point of h, i

Parallelogram ABCD

Parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively also line segments AF and EC intersecting BD

Solution:

Given

  • Parallelogram ABCD

  • E and F are the mid-point of side AB and CD respectively.

To prove, DP=PQ=BQ

Proof,

  • ABCD is a parallelogram

  • Therefore, ABCD and AEFC

Now,

  • AB=CD (Opposite sides of parallelogram ABCD)

  • 12AB=12CD

  • AE=FC (E and F are midpoints of side AB and CD)

  • Therefore, AECF is a parallelogram (AE and CF are parallel and equal to each other)

  • AFEC (Opposite sides of a parallelogram)

Now, In ΔDQC,

  • F is midpoint of side DC and FPCQ (as AFEC ).

  • P is the mid-point of DQ (Converse of mid-point theorem)

  • DP=PQ ……..equation (1)

Similarly, In APB,

  • E is midpoint of side AB and EQAP (as AFEC ).

  • Therefore, Q is the mid-point of PB (Converse of mid-point theorem)

  • PQ=QB …….equation (2)

From equations (1) and (2), DP=PQ=BQ Hence, the line segments AF and EC trisect the diagonal BD.

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

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Quadrilateral poly2 Quadrilateral poly2: Polygon P, S, R, Q Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment e Segment e: Segment [P, S] of Quadrilateral poly2 Segment f Segment f: Segment [S, R] of Quadrilateral poly2 Segment g Segment g: Segment [R, Q] of Quadrilateral poly2 Segment h Segment h: Segment [Q, P] of Quadrilateral poly2 Segment i Segment i: Segment [A, C] Segment j Segment j: Segment [B, D] Segment k Segment k: Segment [P, R] Segment l Segment l: Segment [S, Q] Point A A = (-1.72, 2.82) Point A A = (-1.72, 2.82) Point A A = (-1.72, 2.82) Point B B = (1.6, 3.36) Point B B = (1.6, 3.36) Point B B = (1.6, 3.36) Point C C = (2.78, 0.42) Point C C = (2.78, 0.42) Point C C = (2.78, 0.42) Point D D = (-0.06, -0.38) Point D D = (-0.06, -0.38) Point D D = (-0.06, -0.38) Point P Point P: Point on a Point P Point P: Point on a Point P Point P: Point on a Point S Point S: Point on d Point S Point S: Point on d Point S Point S: Point on d Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point Q Point Q: Point on b Point Q Point Q: Point on b Point Q Point Q: Point on b

Quadrilateral ABCD, PQRS Are Midpoints of Its Sides

Quadrilateral is ABCD and P, Q, R and S are the mid-point of sides AB, BC, CD, DA respectively

Given,

  • ABCD is a quadrilateral

  • P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.

Now, in ΔACD ,

  • R and S are the mid points of CD and DA respectively.

  • Therefore by midpoint theorem, SRAC .

Similarly we can show that,

  • PQAC

  • PSBD

  • QRBD

Thus, PQRS is parallelogram.

Since, PR and QS are the diagonals of the parallelogram PQRS. So, they bisect each other.

Explore Solutions for Mathematics

Sign In