(d) PQ
Using the property that tangents from an external point to a circle are equal:
PA = PD, QA = QB, CS = CB, DS = DR (wait — applying directly):
PA + CS = (PQ – QA) + (QR – QB) ...
Actually: PA + CS = PA + CS. Since PA = PD and QA = QB and CS = CR and DS = DP:
From vertex P: PA = PD; from Q: QA = QB; from R: RB = RC; from S: SC = SD.
PA + CS = PA + CS. Note PQ = PA + AQ and PS = PD + DS = PA + SC = PA + CS.
∴ PA + CS = PS
(c) PS
Source: Chapter 10, Section 10.3 (Theorem 10.2)
From an external point, two tangents are equal. So: PD = PA, SD = SC. Therefore PS = PD + DS = PA + SC = PA + CS. The answer is (c) PS. A common error is confusing which vertices are external points — remember each vertex of the quadrilateral is an external point from which two tangents are drawn to the circle.