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NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 β Part 2
- Mid-point
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it
Q-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.
Solution:
- Give, ABCD is a ABCD is a rectangle and PQRS is a rhombus quadrilateral . P, Q, R
- And S is mid-points of the sides AB, BC, CD and DA respectively.
In
- P and Q are the mid-points of AB and BC respectively
- Thus, and (Mid-point theorem) β¦ equation (1)
In ,
- S, R are the mid-points of AD and DC respectively
- and (Mid-point theorem) β¦ equation (2)
From equation (1) and (2)
- So and
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
- and (Opposite sides of parallelogram) β¦ equation (3)
Now, In ,
- Q and R are mid points of side BC and CD respectively.
- Thus, and (Mid-point theorem) β¦ equation (4)
- (Diagonals of a rectangle are equal) β¦ equation (5)
From equations (1) , (2) , (3) , (4) and (5) ,
- So, PQRS is a rhombus.
Q-4 ABCD is a trapezium in which , 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 Fig.) . Show that F is the mid-point of BC.
Solution:
- Give, ABCD is a trapezium in which ,
- 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
Proof
- F is the mid-point of BC.
BD intersected EF at G.
- In ,
- E is the mid-point of AD and also
- Thus, G is the mid-point of BD (Converse of midpoint theorem)
Now, In ΞBDC,
- G is the mid-point of BD and also GF β₯ AB β₯ DC.
- Thus, F is the mid-point of BC (Converse of midpoint theorem)