1. In the following figure, find the values of x and y and then show that AB || CD. In the figure, we have CD and PQ intersect at Y. ∴ y = 130º [Vertically opposite angles] Again, PQ is a straight line and EA stands on it. ∴ ∠ AEP + ∠ AEQ = 180º [Linear pair] or 50º + x = 180º ⇒ x = 180º - 50º = 130º …(2) From (1) and (2), x = y But they are the angles of a pair of interior alternate angles. ∴ AB || CD. 2. In the following Figure, if AB || CD, CD ||EF and y : z = 3 : 7, find x. ∴ AB || EF and PQ is a transversal. ∴ Interior alternate angles are equal. ∴ ∠ x= ∠ z …(1) Again, AB || CD, ∴ Interior opposite angles are supplementary, ⇒ y + z = 180º But y : z = 3 : 7 3. In the following figure, if AB || CD, EF ⊥ CD and ∠ GED = 126º, find ∠ AGE, ∠ GEF and ∠ FGE. AB || CD and GE is a transversal. ∴ Interior alternate angles are equal. ∴ ∠ AGE = ∠ GED But ∠ GED = 126º [Given] ∴ ∠ AGE = 126º Since ∠ GED = 126º ∴ ∠ GEF + ∠ FED = ∠ GED or ∠ GEF + 90º = 126º or ∠ GEF = 126 ∠ 90° = 36º Next, AB || CD and GE is a transversal. ∴ ∠ FGE + ∠ GED = 180º or ∠ FGE + 126º = 180º or ∠ FGE = 180º - 126º = 54º Thus, ∠ AGE = 126º, ∠ GEF = 36º and ∠ FGE = 54º 4. In the following figure, if PQ || ST, ∠ PQR = 110º and ∠ RST = 130º, find ∠ QRS. Hint: Draw a line parallel to ST through point R. ∵ PQ || ST [Given] and EF || ST [Construction] ∴ PQ || EF and QR is a transversal, ∴ Interior alternate angles are equal i.e ∠ PQR = ∠ QRF But ∠ PQR = 110º [Given] ∴ ∠ QRF = ∠ QRS + ∠ SRF = 110º …(1) Again ST || EF [Construction] and RS is a transversal. ∴ ∠ RST + ∠ SRF = 180º or 130º + ∠ SRF = 180º ⇒ ∠ SRF = 180º - 130º = 50º Now, from (1), we have ∠ QRS + 50º = 110º ⇒ ∠ QRS = 110º - 50º = 60º Thus, ∠ QRS = 60º. 5. In the following figure, if AB || CD, ∠ APQ = 50º and ∠ PRD = 127º, find x and y. We have AB || CD [Given] and PQ is a transversal. ∴ Interior alternate angles are equal. ∴ ∠ APQ = ∠ PQR or 50º = x [∵ APQ = 50º (Given)] ….(1) Again, AB || CD and PR is a transversal. ∴ ∠ APR = ∠ PRD [Interior alternate angles] ⇒ ∠ APR = 127º [∵ It is given that ∠ PRD = 127º] But ∠ APR = ∠ APQ + ∠ QPR ∴ ∠ APQ + ∠ QPR = 127º ⇒ 50º + y = 127º [∵ It is given that ∠ APQ = 50º] ⇒ y = 127º - 50º = 77º Thus, x = 50º and y = 77º 6. In the following figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD. REMEMBER (i) Perpendiculars to the parallel lines are parallel. (ii) According to the laws of reflection, angle of incidence = angle of reflection