
In the given figure, ST is a straight line and ray QP stands on it.

∴ \(∠\)PQS + \(∠\)PQR = 180º (Linear Pair)
\(∠\)PQR = 180º − \(∠\)PQS............ (1)
\(∠\)PRT + \(∠\)PRQ = 180º (Linear Pair)
\(∠\)PRQ = 180º − \(∠\)PRT............ (2)
It is given that \(∠\)PQR =\(∠\) PRQ.
Equating equations (1) and (2), we obtain
180º − \(∠\)PQS = 180º− \(∠\)PRT
\(∠\)PQS = \(∠\)PRT



(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
