Question

Fig. 6.14
In Fig. 6.15, < PQR =
PQS = Z PRT.
PRQ, then prove that
P
S
RT
Fig. 6.15
In Fig. 6.16, if x + y=w+z, then prove that AOB
is a line.​

Answer 1

Linear pair of angles:

If Non common arms of two adjacent angles form a line, then these angles are called linear pair of angles.

Axiom- 1

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°i.e, the sum of the linear pair is 180°.

Axiom-2

If the sum of two adjacent angles is 180° then the two non common arms of the angles form a line.

The two axioms given above together are called the linear pair axioms.

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Solution:

Given,

∠PQR = ∠PRQ

To prove:

∠PQS = ∠PRT

Proof:

∠PQR +∠PQS =180° (by Linear Pair axiom)

∠PQS =180°– ∠PQR — (i)

∠PRQ +∠PRT = 180° (by Linear Pair axiom)

∠PRT = 180° – ∠PRQ

∠PRQ=180°– ∠PQR — (ii)

[∠PQR = ∠PRQ]

From (i) and (ii)

∠PQS = ∠PRT = 180°– ∠PQR

∠PQS = ∠PRT

Hence, ∠PQS = ∠PRT

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Answer 2

Question :-

In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer :-

ST is a straight line.

∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]

Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]

From (1) and (2), we have

∠PQS + ∠PQR = ∠PRT + ∠PRQ

But ∠PQR = ∠PRQ [Given]

∴ ∠PQS = ∠PRT

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