Mathematics Class 9 NCERT Textbook ( 2014) PDF Adithya{ TRRG}

Author:NCERT Textbook (2014) PDF Adithya{TRRG}
Language: eng
Format: epub
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Fig. 8.16

Z DAS + Z ADS =

1 1

-ZA+-ZD

2 2

- (ZA + ZD)

x 180° (Z A and Z D are interior angles

Also, ZDAS + Z ADS + ZDSA:

or, 90° + Z DSA:

or, Z DSA:

So, Z PSR :

90°

180°

180°

90°

90°

on the same side of the transversal) (Angle sum property of a triangle)

(Being vertically opposite to Z DSA)

AIM 1 11 V \ {TRRGr

Quadrilaterals

145

Similarly, it can be shown that Z APB = 90° or Z SPQ = 90° (as it was shown for Z DSA). Similarly, Z PQR = 90° and Z SRQ = 90°.

So, PQRS is a quadrilateral in which all angles are right angles.

Can we conclude that it is a rectangle? Let us examine. We have shown that Z PSR = Z PQR = 90° and Z SPQ = Z SRQ = 90°. So both pairs of opposite angles are equal.

Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and so, PQRS is a rectangle.

8.5 Another Condition for a Quadrilateral to be a Parallelogram

You have studied many properties of a parallelogram in this chapter and you have also verified that if in a quadrilateral any one of those properties is satisfied, then it becomes a parallelogram.

We now study yet another condition which is the least required condition for a quadrilateral to be a parallelogram.

It is stated in the form of a theorem as given below:

Theorem 8.8 : A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

Look at Fig 8.17 in which AB = CD and AB II CD. Let us draw a diagonal AC. You can show that A ABC = A CDA by SAS congruence rule.

So, BC II AD (Why?)

Let us now take an example to apply this property of a parallelogram.

Example 6 : ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and CD (see Fig. 8.18). If AQ intersects DP at S and BQ intersects CP at R, show that:

(i) APCQ is a parallelogram.

(ii) DPBQ is a parallelogram.

(iii) PSQR is a parallelogram.

Fig. 8.18

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